SI, STEM 


A  K     THMETIC, 


THE  P1UACIPLE  OF  CANCKLLNC 


77; 


IN  MEMORIAM 
FLOR1AN  CAJORI 


RECOMMENDATIONS. 


From  Prof.  E.  A,  Andrews. 

About  two  years  ago,  while  on  a  visit  to  this  place,  I  had  the  pleasure  of  meeting 
with  Mr.  Tracy,  and  of  having  some  conversation  with  him  on  the  subject,  of  an 
Arithmetic,  which  he  was  preparing  for  schools.    I  was  particularly  pleased  with 
that  part  of  his  system  which  related  to  canceling,  and  which  appeared  to  me  to  poa-  *  t  *  .  i 
eess  great  practical  value.    Within  a  few  days,  Mr.  Tracy  has  put  into  my  hands  a  1\  I  » 
part  of  his  manuscript,  that  1  might  become  more  minutely  acquainted  with  his  sys- 
tem.   My  time  has  been  so  far  occupied  with  my  own  business,  that  I  have  been  able 
to  examine  a  part  only  of  the  manuscript  put   into  my  hands  ;  but  with  this  part,  I 
have  been  much  gratified.    It  appears  to  me,  that  when  carefully  and  thoroughly  re-  . 
vised  and  perfected,  as  the  author  designs  to  do,  it  may  become  a  most  valuable  work, 
inferior  to  none  of  the  Arithmetics  now  used  in  our  schools.     Such  is  my  confidence  in    '  <> 
the  ability  of  the  author,  to  complete  and  polish  the  work,  that  I  look  upon  its  success 
as  quite  certain. 

New  Britain,  Conn.,  Aug.  1st,  1839. 


From  E,  H.  Burritt,  author  of  the  Geography  of  the  Heavens,  «S-c. 

Through  the  politeness  of  Mr.  Tracy,  I  have  been  favored  with  a  perusal  of  an 
Arithmetic,  in  manuscript,  which  he  is  preparing  for  publication.  The  work  is  in- 
tended as  a  universal  class  book  in  elementary  Arithmetic.  It  is  the  production  of 
a  gentleman  of  known  abilities  and  experience  in  teaching,  and  he  has,  with  great 
care,  arranged  its  several  parts,  and  given  the  rules,  and  selected  the  examples,  step 
by  step,  in  that  natural  order,  and  easy  method,  which  his  own  judgment  and  expe- 
rience approved.  There  are  some  excellencies  in  his  Arithmetic — some  facilities  of 
dealing  with  figures,  which,  so  far  as  I  know,  ai-een^rely  pecujjyr  fn^this)  trfiaHser  91  ft 
and  which  distinguish  it  from  all  others.  On  tfn§^round~especially,  and  that  of  its  "  ' 
"general  nTerlt,  rtKfnTTfa  "work*  which  will  commend  itself  to  the  attention  of 
teachers. 

New  Britain,  August,  1836. 

I  entirely  coincide  in  the  above  opinion,  having  been  particularly  gratified  with  the 
ease  and  facility  with  which  many  difficult  operations  are  performed  by  the  new 
principle  introduced  by  the  author. 

J.  P.  BRACE,  Principal  of  Hartford  Female  Seminary. 


From  Edward  Strong,  Principal  of  Bacon  Academy,  Colchester,  Conn. 

I  have  had  the  pleasure  of  examining  a  system  of  Arithmetic,  by  Mr.  Tracy,  of 
Norwich  Academy.    Without  speaking  of  its  merits  in  other  respects,  which  I  am 
unable  to  do,  from  want  of  time  to  give  a  thorough  perusal,  I  discover  in  it  a  distinc- 
tive feature— the  method  of  canceling— which  appears  to  me  to  be  an  important  im- ||  I* 
provementon  every  other  system  with  which  I  am  acquainted.    Should  the  other  fea-  " 
lures  of  the  work  correspond  with  what  may  reasonably  be  expected  from  its  author, 
I  should  regard  it  as  a  very  important  improvement  upon  our  other  systems  of  Arith- 
metic. 

Colchester,  Oct.  8th,  1839. 


From  Rev.  A.  Bond,  Pastor  of  the  Second  Congregational  Church,  in  Norwich . 

Having  examined  the  general  plan  of  an  Arithmetic,  prepared  by  Mr.  Tracy,  Prin- 
cipal of  Norwich  Academy,  lean  cheerfully  recommend  it  as  a  system  possessing,  in 
some  important  particulars,  a  superiority  over  any  other  system  with  which  I  am 
acquainted.  The  method  of  canceling,  which  is  carried  through  the  work,  excepting 
the  Roots,  greatly  facilitates  the  process  of  Arithmetical  calculations,  and  will  give  it 
a  decided  advantage  in  the  estimation  of  businessmen.  The  part  on  foreign  exchanges 
will  enhance  its  value  with  the  commercial  community.  While  its  simplicity  adapts 
it  to  the  use  of  common  schools,  iis  comprehensiveness,  and  the  ease  and  accuracy 
with  which  complicated  problems  may  be  solved-  will  be  likely  to  secure  for  it  a 
prominent  place  in  the  counting  room. 

Norwich,  August  1th,  1839. 


2  RECOMMENDATIONS. 

From  La  Fayette  S.  Foster,  Esq. 

Mr.  Calvin  Tracy,  of  this  city,  has  submitted  to  my  examination,  in  manuscript, 
an  Arithmetic,  prepared  by  himself,  for  publication.  From  the  known  ability  of  Mr. 
Tracy,  as  an  instructor,  I  was  prepared  to  entertain  a  high  opinion  of  any  treatise 
designed  to  facilitate  the  acquisition  of  knowledge,  of  which  he  might  be  the  author, 
and  from  the  attention  which  [  have  bestowed  on  his  Arithmetic,  I  have  no  hesitation 
in  bearing  testimony  to  its  high  meritorious  character.  His  plan  appears  to  me  to  be 
highly  judicious,  and  ably  and  skillfully  executed.  The  work,  in  my  opinion,  will 
be  a  valuable  addition  to  a  very  important  branch  of  education. 

Norwich,  Dec .  24th,  }  839. 

,  •  7 j  From    Messrs.   J.    H.  Gallup  and    G.  Bushnel,    Teachers  of  the  Eclectic  School, 

Norwich,  Conn. 

Having  examined  Mr.  Tracy's  system  of  Arithmetic,  we  think  it  well  calculated  to 
answer  the  purposes  for  which  it  is  intended.    Mr.  Tracy  has  illustrated,  and  happily 
combined  with  nearly  all  his  operations,  a  method  of  abbreviating  Arithmetical  cal- 
ffr  dilations,  which,  so  far  as  we  know,  has  never  before  been  published  in  any  com- 
fy j  mon  school  Arithmetic.    This  we  deem  an  essential  improvement,  and  are  of  opinion, 
I      that  the  author  has  rendered,  in  this  work,  an  essential  service  to  the  cause  of  edu- 
cation. 
Norwich  City,  Oct.  22<Z,  1839. 


From  Rev.  L.  N.  Tracy,  formerly  Principal  of  New  Britain  Academy, 
I  have  spent  considerable  time  in  a  careful  examination  of  an  Arithmetic  prepared 
by  Mr.  C.  Tracy,  Principal  of  Norwich  Academy.    For  my  own  benefit  and  pleasure, 
1  have  carefully  examined  every  rule,  and  though  I  have  daily  used  the  best  Arilhmetics 
extant,  while  engaged  for  many  years  in  teaching,  I  am  led  to  believe  that  there  is  not  a 
textbook  on  Arithmetic  in  use  which  presents  equal  excellencies.    Its  grand  feature  — 
that  which  distinguishes  it  from  every  other  Arithmetical  treatise  —  is  thcprinciple  of 
canceling,  introduced  and  applied  throughout  the  work.    The  extent  and  facility  of 
*its  application  to  all  operations  in  which  Multiplication  and  Division  are  both  con- 
cerned, are  fully  and  clearly  illustrated.    It  is  safe  to  say  that  two  thirds,  and  often 
four  fifths  of  the  labor  and  time  usually  required  for  arithmetical  solutions,  is  saved. 
While  it  contains  an  amount  of  matter  equal  to  any  other  Arithmetic  in  use,  it  is  still 
a  strictly  elementary  work. 
Norwich,  Oct.  [2th,  1839. 

From  E.  C.  Herrick,  Esq. 

I  have  cursorily  examined  the  manuscript  of  Mr.  C.  Tracy's  treatise  on  Arithmetic. 
The  most  prominent  feature  of  the  work  is  the  introduction  of  a  peculiar  mode  of 
stating  numerous  classes  of  problems,  which  are  then  solved  by  an  abridged  process, 
called  canceling.  This  appears  to  me  an  important  improvement  on  the  books  in 
common  use,  and  one  which  renders  the  publication  of  this  treatise  very  desirable. 

New  Haven,  Dec.  27th,  1839.  % 


From  J.  H.  Rogers,  Esq.,  Principal  of  Prospect  Hill  High  School. 
Messrs.  DURRIE  &  PECK, 

Gentlemen—  From  a  hasty  examination  of  Tracy's  Arithmetic,  I  believe  it  worthy 
of  being  ranked  among  the  best  School  Books.  The  method  of  canceling,  very  fully 
brought  into  practice  in  this  work,  greatly  abridges  many  operations  ;  and  may  be 
mentioned  as  one  of  its  most  valuable  features. 

Sincerely  yours,  J.  H.  ROGERS. 

East  Haven,  April  24th,  1840. 

At  a  meeting  of  School  Visitors  of  the  First  School  Society  of  New  Haven,  held  May 
2,  1840— 

The  committee  appointed  at  the  previous  meeting  to  examine  and  report  on  "A 
New  System  of  Arithmetic,"  by  Mr.  C.Tracy,  —  reported,  that  in  their  opinion  the 
work  contains  important  improvements  on  the  arithmetical  treatises  in  common  use, 
and  recommend  that  it  be  introduced  into  the  schools  of  this  Society:—  whereupon, 
it  was 

footed,  That  Tracy's  New  System  of  Arithmetic  be  adopted  for  use  in  the  schools  of 
the  First  School  Society  of  New  Haven. 

R.  S.  HINMAN,  Chairman. 

E.  C.  HERRICK,  Clerk  pro  tempore. 


NEW    SYSTEM 


OF 


ARITHMETIC, 


IN  WHICH  IS  EXPLAINED  AND  APPLIED  TO  PRACTICAL  PURPOSES,  IN  ADDITION 
TO  THE  ORDINARY  RULES  OF  OPERATION, 


THE  PRINCIPLE  OF   CANCELING; 

BEING  AN 

ABBREVIATED    MODE 

OF 

ARITHMETICAL    SOLUTION. 

DESIGNED    FOR    SCHOOLS    AND    ACADEMIES. 


BY     C.     TRACY,    A.  M 

PRINCIPAL    OF   NORWICH  ACADEMY. 


NEW    HAVEN: 

PUBLISHED    BY    DURRIE    &    PECK. 

NEW  YORK:  COLLINS,  KEESE,  &  co. 

BOSTON  :    CROCKER  &  BREWSTER. 

1840. 


ENTERED, 
According  to  Act  of  Congress,  in  the  year  1840,  by 

DURRIE     &     PECK, 

In  the  Clerk's  Office  of  the  District  Court  of  Connecticut  District. 


HITCHCOCK  AND  STAFFORD,  PRINTERS, 
NEW  HAVKN. 


'& 


INTRODUCTION. 


It  will  readily  be  conceded,  that  all  efforts  in  behalf  of  the  general 
diffusion  of  useful  knowledge,  are  in  themselves  commendable.  There 
is,  however,  and  probably  ever  will  be,  a  difference  of  opinion  rela- 
tive to  the  extent  to  which  books  of  any  particular  description,  and 
treating  upon  the  same  general  topic,  may  be  multiplied,  and  the  inter- 
ests of  education  uniformly  advanced  thereby. 

This  difference  of  opinion  exists  especially  in  relation  to  books  de- 
signed for  the  use  of  common  schools  and  academies,  and  which  treat 
upon  the  more  common  subjects  of  study.  The  multiplication  of 
books  of  this  description,  to  the  extent  realized  at  the  present  day,  is 
regarded  by  many  as  injurious  to  the  general  good.  That  its  tendency 
is  to  increase,  in  some  small  degree,  the  expense  of  education,  at  least 
in  some  parts  of  the  country,  will  not  be  denied.  But  before  sentence 
of  final  condemnation  is  pronounced,  it  always  becomes  those,  who  sit 
as  umpires,  to  take  as  extended  views  of  the  subject  before  them,  as  the 
nature  of  the  case  will  admit. 

The  question  now  presented  is,  how  far  the  general  good  is  advanced 
by  the  multiplication  of  school  books. 

To  answer  this,  let  it  be  supposed  that  only  a  single  work  in  each 
department  of  science  studied  in  common  schools,  had  ever  been  pre- 
sented to  the  public,  aiXl  that  each  work  were  such  as  it  should  be. 
Books  of  this  description  would  obviously  find  ample  circulation,  suf- 
ficient, perhaps,  to  satisfy  both  authors  and  publishers,  without  em- 
bracing one  half  of  the  ground  to  be  occupied.  The  consequence 
would  be,  that  the  more  recently  settled  parts  of  our  country  would  be 
but  poorly  supplied  with  the  means  of  education,  for  at  least  some  con- 
siderable period  of  time.  But  as  it  now  is,  with  such  a  multiplicity  of 
school  books  constantly  emanating  from  the  press,  a  spirit  of  rivalry  is 
created,  a  desire  is  excited  on  the  part  of  both  authors  and  publishers  to 
give  to  their  several  works  a  more  extended  circulation  than  can  be 
obtained  without  exploring  the  whole  ground.  As  a  natural  conse- 
quence,  the  inhabitant  of  the  less  favored  portion  of  our  land,  is  scarcely 
settled  in  his  log  cabin,  before  books  of  every  description  necessary  for 
the  education  of  his  sons  and  daughters,  are  presented  him,  as  it  were, 
at  his  own  door.  His  attention  is  thus  directed  to  a  subject  second  in 
importance  to  none  of  a  temporal  nature  ;  and  one  which,  when  duly 
presented,  will  be  likely  to  be  regarded,  and  to  receive  a  consideration, 
which  otherwise  might  be  long  neglected. 
1* 


e    1  1 
-    t 
' 

r 
,    j  • 


INTRODUCTION. 

The  truth  of  our  supposition,  that  any  one  set  of  school  books  is 
such  in  all  respects  as  is  required,  may,  however,  very  reasonably  be 
doubted.  Many  of  them  are  unquestionably  of  a  very  high  order, 
and  very  probably  owe  some  degree  of  their  merit  to  the  fact,  that 
other  minds  have  been,  are,  and  yet  will  be,  traversing  the  same 
ground  which  their  authors  trod,  and  are  preparing  other  works,  to 
supersede  them,  if  possible,  in  the  estimation  of  the  public.  t 

^       The  effect  of  the  multiplication  of  school  books  is,  therefore,  to  ren- 
!    der  the  means  of  education  as  perfect  as  the  nature  of  their  subjects  will 
,     allow,  and  to  convey  these  means,  thus  perfected,  to  every  part  of  our    \ 
*    entire  country. 

From  the  preceding  considerations,  the  author  is  inclined  to  regard 
the  multiplication  of  school  books  as  favorable  to  the  cause  of  general 
education.  It  therefore  only  remains  to  point  out  some  of  the  more 
important  features  of  the  following  work,  before  introducing  it  to  the 
ordeal  of  public  opinion.  That  it  is  worthy  of  public  attention  and 
patronage,  belongs  not  to  him  to  decide.  4t  certainly  will  be  found  to 
possess  some  peculiarities,  which  are  of  course  regarded  by  him  as  im- 
provements. Whether  they  are  indeed  such,  remains  for  others  to  de- 
teimine. 

A  peculiar  feature  of  the  following  pages,  and  one  which  distin- 
guishes this  work  from  every  other  on  the  same  subject,  is  the  "  Sys- 
tem of  Canceling,"  which,  in  connection  with  the  ordinary  mode  of 
solution,  is  introduced  throughout,  and  applied  to  such  arithmetical 
problems  as  embrace  in  their  operation  both  multiplication  and  divis- 

:/  ion.  This  is  regarded  by  the  author,  and  by  many  others  acquainted 
I  with  his  system,  as  a  decided  improvement  upon  all  Arithmetics  here- 
1  tof ore  presented  to  the  public. 

The  following  are  some  of  the  advantages  of  the  new  system  : 
I       1st.  The  statement  required,  or,  rather,  recommended  for  canceling, 
is  itself  a   complete  analysis  of  the  sum  proposed.    Suppose,  for  illus- 
tration, that  12  yards  of  cloth  cost  $48,  and  that  it  is  required  to  find 
the  value  of  15  yards  of  the  same. 

.We  analyze  the  preceding  sum,  either  by  first  finding  the  value  of 
one  yard  of  the  cloth,  viz.  $48-*- 12  yd.  =  $4,  and  then  multiplying 
that  price  by  the  number  of  yards,  as,  $4X15  yd.  =  $60,  Ans. ;  o"r  by 
finding  the  ratio  of  the  number  of  yards  of  which  the  price  is  given, 
and  of  those  of  which  the  price  is  required,  and  then  multiplying 
that  ratio  by  the  given  cost.  This  ratio  is  -J-f  =f- ;  and  Jx48  =  60,  the 
number  of  dollars  required,  and  the  same  as  above. 
The  same  stated  for  canceling: 

48.15 
12' 

.     (See  rule  for  canceling,  in  Single  Proportion.)    By  the  above  state- 
'  ment,  $48  is  to  be  increased  by  the  ratio,  12 : 15 ;  or  \^  =  -|.    It  is,     . 
however,  obvious  that-jL  of  48  multiplied  by  15,  is  the  s"ame  as  ^  of 
15  multiplied  by  48.    The  above  sum  is  therefore  canceled  and  solved   A 
*  »  thus:  I 

4  if 

48,    15 
— and  15X4  =  60,  the  dollars  required. 


INTRODUCTION.  7 

The  application  of  the  canceling  principle  is,  however,  more  com- 
pletely illustrated  by  the  solution  of  sums  in  which  the  ratio  of  one  of 
the  given  quantities  to  a  required  quantity  of  the  same  kind,  is  traced 
through  several  simple  ratios.  The  following  sum  may  serve  as  an 
illustration  : 

If  3  men,  in  16  days,  of  9  hours  each,  build  a  wall  20  feet  long,  6 
feet  high,  and  4  feet  thick,  in  how  many  days,  of  8  hours  each,  will  12 
men  build  a  wall  200  feet  long,  8  feet  high,  and  6  feet  thick  1 

It  is  obvious  the  ratio  of  the  given  days  to  the  required  number  of 
days,  is  compounded  of  the  ratios,  8  hours  :  9  hours;  12  men  :  3  men; 
20  feet  in  length  :  200  feet  in  length  ;  6  feet  in  height :  8  feet  in  height; 
and  4  feet  in  thickness :  6  feet  in  thickness.  Or,  these  ratios  may  be 
fractionally  expressed,  thus  :  -f,  T32,  2^°,  f ,  and  -f . 

Now  the  given  days  of  9  hours  each  are  changed  to  days  of  8  hours 
each,  by  the  following  statement : 

^  =  18  day, 

The  time,  in  days  of  8  hours  each,  required  for  the  12  men  to  com- 
plete the  work  of  3  men,  is  obtained  by  uniting  the  second  of  the  pre- 
ceding ratios  to  the  above  statement ;  thus  : 

16.  9.    3 


By  introducing  into  the  same  statement,  the  third  of  the  preceding 
ratios,  we  obtain  the  requisite  lime  for  completing  the  200  feet  of  wall, 
allowing  the  height  and  thickness  of  each  wall  to  be  the  same ;  thus : 

16.  9.    3.  200 

If  the  fourth  ratio  be  united,  we  obtain  the  time  required,  allowing 
each  wall  to  be  of  the  same  thickness  ;  thus  : 

16.  9.    3.  200.  8 
8.  12.    20.  6 

Lastly,  if  the  fifth  and  last  ratio  be  introduced,  the  number  of  days 
required  by  all  the  conditions  of  the  question,  is  obtained  ;  viz. 

16.  9.    3.  200.  8.  6 
8.  12.    20.  6.  4 
The  last  statement  canceled : 
10   S 

1S  I  12   *Stt'  ft   4'  and  10X9  =  9°  days>  answer- 
lS 

2dly.  A  second  advantage  to  be  derived  from  the  canceling  system,  is  ll 
the  facility  afforded  by  it  for  reducing  several  operations  to  a  single  I  ( 
statement.  The  following  example  will  afford  an  illustration  : 

Bought  742  Ib.  of  wool,  a  deduction  of  5  per  cent,  from  the  gross 
weight  being  made,  for  dust,  &c.  For  the  net  weight,  I  paid  9  s.  New 
York  currency,  per  Ib.,  and  for  ready  money,  was  allowed  a  deduction 


8  INTRODUCTION. 

of  6  per  cent.    Now,  allowing  that  I  sold  the  same  so  as  to  realize  a 
gain  of  20  per  cent.,  how  much  money  did  I  receive  1 

The  ordinary  mode  of  solving  this  sum  requires  the  four  following 
operations;  viz.  105  :  100  :  :  742  :  the  net  weight  of  the  wool,  which  is 
706|  Ib.  Again,  706f  X9-V-8  =  795,  the  number  of  dollars  which  the 
wool  would  have  cost,  if  no  deduction  had  been  made  for  ready  money. 
But  a  deduction  of  6  per  cent,  was  actually  made  ;  therefore,  106  :  100  :  : 
795  :  the  money  paid,  viz.  $750.  Now,  on  this  last  sum,  I  realized  a 
gain  of  20  per  cent.  ;  hence,  100  :  120:  :  750  :  the  money  received,  viz. 
$900,  Ans.  By  canceling,  these  four  statements  are  reduced  to  one, 
thus: 

742.  100.  9.  100.  120 
105.  8.  1067100' 

that  is,  the  742  Ib.  is  to  be  multiplied  by  the  four  succeeding  ratios,  and 
the  number  obtained  will  equal  the  number  of  dollars  required. 

The  same  canceled  : 

7     20    3  15 


11 

s 

3dly.  A  great  advantage  of  the  canceling  system  over  all  others,  arises 
from  the  expedition  it  affords  in  arithmetical  solutions.  Instead  of 
multiplying  and  dividing  by  all  the  numbers  which  the  nature  of  the 
sum  proposed  would  naturally  require,  the  multipliers  and  divisors  are 
made  to  cancel  each  other ;  that  is,  equal  factors  are  rejected  from  both. 
Hence,  they  are  all  made  to  exert  their  appropriate  influence  in  pro- 
curing the  answer,  while  the  labor  of  multiplying  and  dividing  is 
avoided.  The  statement  of  each  sum  for  canceling  is  a  fractional 
answer  of  the  same,  and  it  is  obvious,  that  the  value  of  fractions  is  not 
affected  by  rejecting  equal  factors  from  their  numerators  and  denomi- 
nators. 

The  processes  of  reduction  which  occur  very  frequently  in  common 
Arithmetics,  are  mostly  avoided  by  this  system.  Suppose  it  be  required 
to  find  how  many  pounds  sterling  5  hogsheads  of  wine  would  cost 
at  10  d.  per  pint.  By  the  canceling  system,  it  is  necessary  only  to 
write  down  the  numbers  required  to  effect  the  reduction,  and  the  ques- 
tion is  then  solved  by  canceling  those  numbers  as  far  as  practicable. 
Thus: 

5.  63.  4.  2.  10 
"  12.  20' 

The  numbers  above  the  line  are  obviously  those,  which,  when  mul- 
tiplied together,  will  give  the  answer  in  pence.    The  numbers  below 
the  line,  are  those  required  to  reduce  pence  to  pounds  sterling.     The 
above  sum  canceled : 
21 


i  then>  21X5  =  105  £'  Ans' 


That  the  above  method  of  solving  arithmetical  problems  is  easily 


INTRODUCTION.  9 

comprehended  and  applied  by  the  scholar,  has  been  fully  tested  by  the 
author.  The  experience  of  nine  or  ten  years  entirely  devoted  to  the 
business  of  instruction,  leaves  him  no  room  to  doubt  on  this  point. 
Being,  however,  fully  aware  that  his  Arithmetic  might  fall  into  the 
hands  of  some,  who  would  not  at  once  comprehend  and  apply  the  prin- 
ciple of  canceling,  he  has  introduced  the  ordinary  rules  of  solution,  in 
connection  with  those  of  canceling,  and  has  endeavored  to  render 
both  modes  plain  and  familiar,  by  frequent  and  clear  illustrations. 

The  constant  aim  of  the  teacher  should  be  to  prepare  his  pupils  for 
the  active  duties  of  life  ;  and,  in  the  department  of  Arithmetic,  this  is 
accomplished  only  when  the  scholar  has  acquired  correctness  and  ex- 
pedition in  effecting  his  solutions. 

To  make  good  arithmeticians,  it  is  first  necessary  to  acquire  a  correct 
and  extensive  comprehension  of  the  simple  or  fundamental  rules  of 
Arithmetic.  When  this  is  done,  their  application  will  be  obvious.  The 
danger,  therefore,  is  not,  that  the  scholar  will  spend  too  much  time  on 
what  is  usually  regarded  as  the  more  simple  part  of  Arithmetic,  but 
that  he  will  leave  it  too  soon. 

In  the  use  of  this  treatise,  the  author  would  recommend,  that,  when 
the  pupil  shall  have  passed  the  simple  rules,  and  commenced  those  ope- 
rations to  which  canceling  may  be  applied,  he  be  required  to  solve 
each  problem  both  by  the  ordinary  rule,  and  by  the  rule  for  canceling. 
More  practice  will  thus  be  secured,  and,  consequently,  greater  expedi- 
tion acquired. 

In  the  illustrations,  which  are  given  in  connection  with  the  different 
rules,  it  has  been  the  design  of  the  author  fully  to  acquaint  the  scholar 
with  the  nature  of  the  subject  presented,  without  carrying  his  expla- 
nations so  far  as  to  take  the  work  which  properly  belongs  to  the 
scholar,  out  of  his  hands.  No  important  acquisition  can  be  made, 
without  corresponding  effort.  This  fact  seems  to  have  been  overlooked, 
in  the  preparation  of  some  Arithmetics  now  in  use,  and  special  effort 
made  to  render  every  thing  as  easy  as  possible  for  the  scholar ;  that  is, 
to  enable  him  to  effect  the  solutions  with  very  little  mental  labor.  The 
intellectual  powers  are,  however,  developed  and  strengthened  only  by 
being  brought  into  vigorous  exercise.  "  In  Arithmetic,  the  young  be- 
ginner should  find  just  enough  assistance  to  encourage  and  stimulate 
him  to  effort.  That  is  not  the  best  system,  which  enables  the  learner 
to  advance  from  rule  to  rule  with  the  least  amount  of  study;  but  that, 
which,  while  it  helps  him  over  some  difficulties,  leaves  him  examples 
V  enough  to  task  his  powers  to  the  utmost."  (Dr.  Humphrey's  Thoughts 
on  Education.)  With  these  introductory  remarks,  the  following  work 
is  commended  to  the  candor  of  an  enlightened  public. 

THE  AUTHOR. 

Norwich  Academy,  May,  1840. 


Ill 


ERRATA. 

Notwithstanding  care  has  been  constantly  exercised  to  avoid 
errors,  the  following  will  be  found  to  have  escaped  detection  : 

Page  42,  line  25,  read  G,  instead  of  5  hundred. 
"      65,  in  the  4th  sum,  and  in  the  first  statement  of  that  sum,  read 

21,  instead  of  4. 

"     119,  line  34,  read  |,  instead  of  f . 
"    136,  line  14,  read  integers,  for  integris. 
"    171,  line  7,  read  $420,  instead  of  $4.20. 


TABLE   OF   CONTENTS. 


PAGE 

Arithmetic, 13 

Notation, 

Numeration, 14 

Simple  Addition, 

Simple  Subtraction,      ......... 

Simple  Multiplication, 

Simple  Division, 43 

Federal  Money,        .                 54 

Addition  of  Federal  Money, 56 

Subtraction  of  Federal  Money 57 

Multiplication  of  Federal  Money, 58 

Division  of  Federal  Money, 

Canceling,             61 

Compound  Numbers,        ........  67 

Reduction  of  Compound  Numbers, 72 

Addition  of  Compound  Numbers,            93 

Subtraction  of  Compound  Numbers,            101 

Multiplication  of  Compound  Numbers,           ....  107 

Division  of  Compound  Numbers, 112 

Vulgar  Fractions, 117 

Reduction  of  Fractions, 123 

To  reduce  whole  or  mixed  numbers  to  Improper  Fractions,  123 
To  reduce  Improper  Fractions  to  whole  or  mixed  numbers,      ;  124 
To  reduce  Compound  Fractions  to  Simple  ones,     . 
To  change  Fractions  from  one  denomination  to  another,            .  126 
To  find  the  integral  value  of  Fractions,           ....  129 
To  reduce  low  denominations  to  Fractions  of  higher  denomina- 
tions,             130 

To  reduce  Fractions  to  a  Common  Denominator,   .        .        .  131 

Addition  of  Fractions, 132 

Subtraction  of  Fractions, 

Multiplication  of  Fractions,        .......  135 

Division  of  Fractions, 

Decimal  Fractions, 142 

Addition  of  Decimal  Fractions, 146 

Subtraction  of  Decimal  Fractions, 147 

Multiplication  of  Decimal  Fractions, 

Division  of  Decimal  Fractions, 149 

Reduction  of  Vulgar  and  Decimal  Fractions,         .        .        •  152 
Reduction  of  Decimal  to  Vulgar  Fractions,        ....  152 
Reduction  of  low  denominations  to  Decimals  of  a  higher  denom- 
ination,             153 


12  CONTENTS. 

PAGE 

To  find  the  integral  value  of  a  Decimal,         ....  155 

Reduction  of  Currencies,             157 

Simple  Interest, 162 

Banking, 174 

Compound  Interest, 179 

Commission, 181 

Insurance, 181 

Ratio, 183 

Proportion, 185 

Compound  Proportion, 196 

Analytical  Solution,        .                                                  .  202 

Simple  and  Compound  Proportion  in  Fractions,        .        .        .  206 

Conjoined  Proportion, 209 

Discount,      .                211 

Profit  and  Loss, 213 

Barter, 217 

Partnership, 219 

Commercial  Exchange, 223 

Table  of  Gold  Coins 

Table  of  Uncoined  and  Silver  Money, 226 

Tare  and  Tret, 230 

Equation  of  Payments, 

Duodecimals, 235 

Involution, 239 

Evolution, 241 

Extraction  of  the  Square  Root, 243 

Extraction  of  the  Cube  Root, 

Arithmetical  Progression,            261 

Geometrical  Progression, 264 

Alligation,            268 

Position, 

Double  Position, 

Promiscuous  Examples, 

APPENDIX, 284 


ARITHMETIC. 


ARITHMETIC  explains  the  properties  and  relation  of  num- 
bers, and  makes  known  their  practical  application. 

There  are  six  fundamental  operations,  with  which  the 
scholar  must  become  perfectly  familiar  before  he  can  advance 
successfully ;  viz.  Notation,  Numeration,  Addition,  Subtrac- 
tion, Multiplication,  and  Division.  These  operations  are  call- 
ed fundamental,  because  all  others  are  founded  upon,  or  are 
wrought  by  the  application  of  one  or  more  of  them.  They 
therefore  require  to  be  first  clearly  understood. 

NOTATION. 

Notation  is  the  art  of  expressing  numbers  by  numerical 
characters.  The  characters  employed  to  express  numbers 
are,  1,  2,  3,  4,  5,  6,  7,  8,  9,  0,  and  are  called  figures.  Each 
of  these  figures  has  its  own  specific,  and  also  its  local  value, 
as  will  be  learned  from  Numeration.  Besides  these  charac- 
ters, there  are  others  used  to  express  operations. 

1st.  The  sign  of  addition,  viz.  +  >  (or  plus,  more  ;)  requiring 
the  numbers  between  which  it  is  placed  to  be  added :  3  +  2 
are  5  ;  that  is,  3  added  to  2  are  5  ;  usually  read,  3  plus  2  are  5. 

2d.  The  sign  of  subtraction,  viz.  — ,  (or  minus ;)  showing 
that  the  number  following  is  to  be  taken  from  that  which  pre- 
cedes it ;  thus,  4  —  2  is  2  ;  that  is,  2  taken  from  4,  2  remains. 

3d.  The  sign  of  multiplication,  viz.  x  ;  requiring  the  num- 
ber placed  before  it  to  be  multiplied  by  that  which  follows ; 
thus,  3  X  4  is  12  ;  that  is,  3  multiplied  by  4  is  12. 

4th.  The  sign  of  division,  viz.  H-;  requiring  the  number 
preceding  it  to  be  divided  by  that  which  follows  ;  thus,  8-i-2 
is  4  ;  that  is,  8  divided  by  2  is  4. 

In  the  use  of  each  of  the  preceding  signs,  the  figure  prece- 
ding the  sign  is  to  be  operated  upon  by  that  which  follows  it. 
2 


14  NUMERATION. 

5th.  The  signs  of  proportion,  viz.  :  ::  : ;  showing  that  the 
numbers  including  and  between  these  dots,  are  proportionals ; 
thus,  2  :  4  ::  6  :  12  ;  that  is,  2  bears  the  same  relation  to  4  as 
6  to  12.  The  numbers  are  thus  read,  2  is  to  4  as  6  is  to  12. 

6th.  The  sign  of  equality,  viz.  =;  expressing  the  equality 
of  the  numbers  between  which  it  is  placed ;  or  that  the  num- 
bers on  the  right  equal  those  on  the  left ;  thus,  9  +  7=20  —  4. 

7th.  The  characters  V,  V,  V,  V,  &c.  require  some  root  of 
the  number  before  which  they  stand  to  be  extracted.  The 
figure  placed  over  the  sign  always  shows  what  root  is  required. 
When  the  character  is  used  without  any  figure,  it  then  Indi- 
cates the  square  root. 

By  the  use  of  these  characters,  any  arithmetical  operation 
may  be  indicated.  If  it  be  required  to  add  9  to  16,  from  the 
amount  to  subtract  5,  to  divide  the  remainder  by  4,  and  to  mul- 
tiply the  quotient  by  6,  the  operation  would  be  thus  expressed : 
9+16  — 5-^4x6  =  30. 

QUESTIONS. — What  does  Arithmetic  explain  1  What  application 
does  it  make  of  numbers  1  How  many  are  the  fundamental  operations 
of  Arithmetic  1  What  are  they  ?  Why  are  these  called  fundamental 
operations  1  What  is  Notation?  What  are  the  characters  used  to 
express  numbers  called  *?  What  two-fold  value  has  each  figure  ? 
For  what  purposes  are  other  characters  used*?  What  is  the  sign  of 
addition,  and  for  what  is  it  used  1  The  sign  of  subtraction  1  What 
does  it  require  1  The  sign  of  multiplication  1  What  does  it  require  1 
The  sign  of  division  1  What  does  it  require  1  The  signs  of  propor- 
tion 1  What  do  they  show  ?  The  sign  of  equality  1  What  does  it 
show  ?  What  is  the  character  used  to  express  the  extraction  of  roots 
called1?  Ans.  The  radical  sign.  What  does  the  figure  placed  over 
the  radical  sign  show  7 


NUMERATION. 

The  scholar  has  seen,  under  Notation,  the  characters  used 
to  express  the  first  nine  numbers,  viz. ;  that  to  express  one  whole 
object  or  thing,  1  is  used ;  to  express  two  whole  things,  2  is 
employed  ;  and  for  three  whole  things,  3  is  taken,  &c.,  so  that 
each  character  has  its  own  specific  value  ;  and  this  it  always 
expresses  when  it  stands  alone.  But  each  figure  has  also  a 
local  value,  that  is,  a  value  depending  on  the  place  it  occupies  ; 


NUMERATION.  15 

thus,  the  value  of  3  differs  in  each  of  the  following  numbers, 
viz.  003,  030,  and  300.  In  the  first  number  its  value  is  three 
units,  or  ones  ;  in  the  second  number  it  is  three  tens,  or  thirty  ; 
and  in  the  last,  it  is  three  hundreds.  It  will  therefore  be 
readily  perceived,  that  the  position  of  a  figure  materially  affects 
its  value.  Numeration  teaches  how  to  determine  what  this 
value  is ;  and  thus  it  also  enables  us  to  determine  the  total 
value  of  any  number  of  figures.  It  will  be  found  that  any 
figure  is  increased  in  a  ten-fold  ratio  by  having  a  single  figure 
placed  on  the  right  of  it ;  thus,  6  alone,  is  6  units ;  but  if 
another  figure  be  placed  on  the  right  of  this,  its  value  is  ten 
times  as  great  as  before  ;  thus,  in  63,  the  6  is  six  tens,  equal  to 
60,  and  the  3  is  three  units.  This  value  is  increased  a  hun- 
dred-fold by  having  two  figures  placed  on  the  right ;  thus,  600  ; 
and  a  thousand-fold  by  having  three  figures  on  the  right  of  it ; 
thus,  6000.  In  the  first  of  the  last  two  examples,  the  value  of 
6  is  six  hundred,  and  in  the  second  it  is  six  thousand.  Hence, 
the  scholar  will  see  the  necessity  of  terms  by  which  to  desig- 
nate this  local  value  of  figures,  and  will  also  readily  see  the 
appropriateness  of  those  used,  viz.  Units,  Tens,  Hundreds, 
Thousands,  Tens  of  Thousands,  Hundreds  of  Thousands, 
Millions,  Tens  of  Millions,  Hundreds  of  Millions,  &c.  These 
nine  terms  are  sufficient  to  express  any  number  in  common 
practice.  The  higher  denominations  are  Billions,  Tens  of 
Billions,  Hundreds  of  Billions  ;  Trillions,  Tens  of  Trillions, 
Hundreds  of  Trillions ;  Quadrillions,  Tens  of  Quadrillions, 
Hundreds  of  Quadrillions  ;  Quintillions,  Tens  of — Hundreds 
of — ;  Sextillions,  Tens  of — Hundreds  of — ;  Septillions,  Tens 
of — Hundreds  of — ;  Octillions,  Tens  of — Hundreds  of — ; 
Nonillions,  Tens  of — Hundreds  of — j  &c. 

It  will  be  observed  that  as  the  first  three  figures,  reckoning 
from  the  right,  are  units,  tens,  and  hundreds,  so  every  suc- 
ceeding three  are  appropriated  to  the  units,  tens,  and  hun- 
dreds of  the  succeeding  higher  denominations.  The  following 
table  will  serve  as  an  illustration : 


369,  342,  900,  976,  368,  265,  371,  502,  634,  436. 

3  gg     -ss    -8  S3    -SS3    -5S2    -§£.3    -SS.3    -5  S3    -5  S3    -S  S  3 

£2  =      22=     £2=     £2=     £2=     £2=      £S=      £2 c     £2«     £2g 

•O        3-c        3-0        3-0        3T33>a        3-0        3-C        3TJ        3-O        3 

SCCCCCCCCB 

3333333339 

XSESESSEWS 


16  NUMERATION. 

This  table  will  enable  the  scholar  to  see  at  a  glance,  that 
the  names  and  value  of  figures  are  entirely  dependent  on4heir 
location.  If  they  be  counted  from  the  right  hand  towards  the 
left,  the  first  figure  in  any  line  of  figures  is  units  ;  the  second 
is  tens  ;  the  third,  hundreds  ;  the  fourth,  thousands  ;  the  fifth, 
tens  of  thousands,  &c. ;  and  whatever  station  or  place  any 
figure  may  occupy,  its  value  becomes  ten  times  as  great  by 
being  moved  one  degree  farther  to  the  left. 

Q.UESTIONS. — What  are  the  characters  used  to  express  the  first  nine 
numbers'?  Give  an  example.  When  does  each  figure  express  its 
own  specific  value?  What  other  value  has  each  figure ?  On  what 
does  the  local  value  of  a  figure  depend  1  Give  an  illustration.  What, 
therefore,  does  Numeration  teach  us  to  do  1  What  does  it  enable  us 
to  do  1  In  what  ratio  is  the  value  of  any  figure  increased  by  having  a 
single  figure  placed  on  the  right  7  In  what  ratio  is  it  increased  "by 
having  two  figures  on  the  right  of  it  1  In  what  by  having  three  placed 
on  its  right  1  In  what  ratio  do  numbers  continue  to  increase  from  the 
right  to  the  left  1  Ans.  In  a  ten-fold  ratio.  What  are  the  terms  by 
which  the  local  value  of  figures  are  expressed  1  To  what  are  each 
three  successive  figures  in  any  number  appropriated  1  Ans.  To  the 
units,  tens,  and  hundreds  of  each  denomination.  In  tracing  the  figures 
from  the  right  to  the  left,  what  is  the  first  figure  called  1  The  second  7 
The  third  1  The  fourth  ?  &c.  What  effect  is  produced  on  the  value 
of  any  figure  by  moving  it  one  place  to  the  left  7 

Enumerate  the  following  numbers,  viz.  6  ;  27  ;  467  ;  568  ; 
4269;  13786;  27599;  367595;  17129567;  67596422;  586; 
379872689 ;  278063  ;  596402606;  295  ;  336003;  300300303; 
505050505; 467327986427; 585950876; 688001 ; 100000000; 
99999999999;  6398742913;  4678. 

The  scholar  should  be  taught  to  read  these  figures  accurate- 
ly ;  for  example,  suppose  he  be  required  to  enumerate  the  last 
number,  viz.  4678  ;  let  him  commence  and  repeat  thus  :  eight 
units,  seven  tens,  six  hundreds,  and  four  thousands  ;  and  then 
unite  them,  thus  :  four  thousand  six  hundred  and  seventy-eight. 
Let  him  also  be  required  to  give  the  value  of  any  figure  as  it 
may  vary  by  being  written  at  different  points  under  any  line  of 
figures. 

After  the  scholar  has  become  familiar  with  the  preceding 
exercise,  he  may  write  the  following  numbers  on  his  slate  in 
figures,  taking  care  to  express  each  number  accurately : — 
1.  Thirty-five.  2.  Three  hundred  and  seventy-five.  3.  Three 
hundred  and  five.  4.  Seven  thousand  six  hundred  and  thirty- 
five.  5.  Seven  thousand  and  thirty-five.  6.  Seventy-five 
thousand  four  hundred  and  sixteen.  7.  Seventy-five  thousand 
and  sixteen.  8.  Seventy-five  thousand  and  six.  9.  Seventy- 


ADDITION  OF  SIMPLE  NUMBERS. 


17 


five  thousand.  10.  Three  hundred  and  thirty-three  thousand 
three  hundred  and  thirty-three.  11.  Three  hundred  thousand 
and  three.  12.  Three  hundred  thousand  three  hundred  and 
three.  13.  Five  millions  and  five.  Six  millions  and  seventy- 
five.  One  hundred  and  sixty  millions.  Forty-seven  millions, 
one  hundred  and  five  thousand  and  sixty.  14.  Qft^  hundred 
millions,  one  hundred  and  one.  15.  One  hundreoTahd  seven 
millions,  one  hundred  and  seven  thousand,  one  hundred  and 
seven.  16.  Two  billions,  three  hundred  and  three  millions,  five 
hundred  and  five  thousand  and  six.  17.  Seven  hundred  and 
seven  trillions,  six  hundred  and  seventy -two  billions,  nine  mill- 
ions, three  hundred  and  five  thousand,  six  hundred  and  nine. 

There  is  yet  another  method  of  expressing  numbers  ;  viz.  the 
Roma*n  method  ;  in  which  the  letters  of  the  alphabet  are  used, 
as  may  be  seen  from  the  following  table, 


ROMAN  TABLE. 


I One. 

II Two. 

Ill Three. 

IV Four. 

V Five. 

VI Six. 

VII Seven. 

VIII Eio-ht. 

IX Nine. 

X Ten. 

XX Twenty. 

XXX Thirty. 

XL Forty. 

L Fifty. 


LX  . 
LXX. 
LXXX 
XC  . 

C 

CC  . 
CCC  . 

cccc 

D 
DC     . 

DCC. 
DCCC 
DCCCC 
M 


Sixty. 
Seventy. 
Eighty. 
Ninety. 
One  hundred. 
Two  hundred. 
Three  hundred. 
Four  hundred. 
Five  hundred. 
Six  hundred. 
Seven  hundred. 
Eight  hundred. 
Nine  hundred. 
One  thousand. 


ADDITION   OF    SIMPLE   NUMBERS. 

It  is  highly  important  that  the  scholar  obtain  clear  and  dis- 
tinct views  of  the  nature  of  what  he  has  to  perform.     He  is 
therefore  recommended  to  make  himself  familiar  with  the  fol- 
2* 


18 


ADDITION  OF  SIMPLE  NUMBERS. 


lowing  table  in  the  first  place  ;  then  to  take  the  mental  exer- 
cises that  follow,  and  study  them  faithfully  before  com- 
mencing with  a  slate.  Indeed,  the  author  would  advise,  that 
the  tables,  together  with  the  mental  exercises,  which  stand  at 
the  beginning  of  each  of  the  four  simple  rules,  be  first  studied  ; 
and  that  the  beginner  then  turn  back  and  review  the  whole. 


ADDITION  TABLE. 

The  signs  plus  and  minus,  &c.  are  introduced  into  the  fol- 
lowing tables,  that  the  scholar  may  early  learn  the  use  of  them. 
He  will  therefore  turn  back  to  Notation,  if  he  does  not  recol- 
lect their  use. 


2  plus  1  equals  3 

3+1=4 

4+1=5 

5+1=6 

2+2=4 

3+    2=    5 

4+2=6 

5+2=7 

2+3=5 

3+3=6 

4+3=7 

5+3=8 

2+    4  =     6 

3+4=7 

4+4=8 

5+4=9 

2+5=7 

3+5=8 

4+5=9 

5+    5  =  10 

2+6=8 

3+6=9 

4+    6  =  10 

5+    6  =  11 

2+    7  =     9 

3+    7=  10 

4+    7=  11 

5+    7=  12 

2+    8  =  10 

3+    8  =  11 

4+    8  =  12 

5+    8  =  13 

2+    9  =  11 

3+    9  =  12 

4+    9  =  13 

5+    9  =  14 

2  +  10  =  12 

3  +  10  =  13 

4+  10  =  14 

5+10  =  15 

2  +  11  =  13 

3+11  =  14 

4  +  11  =  15 

5  +  11  =  16 

2  +  12  =  14 

3+12  =  15 

4+12  =  16 

5  +  12  =  17 

6+1=7 

7+1=8 

8+1=9 

9+    1  =  10 

6+2=8 

7+2=9 

8+    2  =  10 

9+    2  =  11 

6+3=9 

7+    3  =  10 

8+    3  =  11 

9+    3  =  12 

6+    4  =  10 

7+    4  =  11 

8+    4=  12 

9+    4  =  13 

6+    5  =  11 

7+    5  =  12 

8+    5  =  13 

9+    5  =  14 

6+    6  =  12 

7+    6  =  13 

8+    6  =  14 

9+    6=  15 

6+    7=  13 

7+    7=  14 

8+    7  =  15 

9+    7  =  16 

6+    8  =  14 

7+    8  =  15 

8+    8  =  16 

9+    8=  17 

6+    9  =  15 

7+    9=  16 

8+    9  =  17 

9+    9  =  18 

6+10=  16 

7  +  10  =  17 

8+10  =  18 

9+10=  19 

6  +  11  =  17 

7+11  =  18 

8+11  =  19 

9+11  =  20 

6  +  12  =  18 

7  +  12  =  19 

8+12  =  20 

9  +  12  =  21 

ADDITION  OF  SIMPLE  NUMBERS. 


19 


10+  1  =  11 

11  +  1  =  12 

12  +  1  =  13 

10+  2  =-12 

11  +  2  =  13 

12  +  2  =  14 

10+  3  =  13 

11  +  3  =  14 

12  +  3  =  15 

10+  4  =  14 

11  +  4  =  15 

12  +  4  =  16 

10+  5  =  15 

11  +  5  =  16 

]2  +  5  =  17 

10  +  6  =  16 

11  +  6  =  17 

12  +  6  =  18 

10+  7=  17 

11  +  7  =  18 

12+  7  =  19 

10  +  8  =  18 

11  +  8  =  19 

12  +  8  =  20 

10+  9  =  19 

11+  9  =  20 

12+  9  =  21 

10+  10  =  20 

11  +  10  =  21 

12  +  10  =  22 

'10+  11  =21 

11  +  11  =  22 

12  +  11  =23 

10  +  12  =  22 

11  +  12  =  23 

12  +  12  =  24 

In  reciting  this  and  the  following  tables,  the  teacher  should 
be  careful  that  the  scholar,  as  he  repeats  his  answers,  per- 
form the  necessary  mental  operation ;  he  must  teach  his  pupil 
to  think. 

MENTAL  EXERCISES  IN  ADDITION. 

If  you  borrow  6  dollars  at  one  time  and  3  at  another,  how 
many  will  you  have  ?  6  and  3  are  how  many  ?  A  pays 
you  3  dollars,  B  4,  and  C  5  ;  how  many  do  they  all  pay  you  ? 

3  +  4+5  =  how  many  ?     If  you  pay  away  6  cents  at  one  time, 

4  at  another,  and  3  at  another,  how  many  do  you  pay  away  ? 
6  +  4  +  3  =  how  many  ?     If  you  buy  twelve  apples  for  9 
cents,  and  a  pint  of  chesnuts  for  4,  how  many  cents  do  the 
apples  and  chesnuts  cost  ?     9  +  4  =  how  many  ?     John  gave 
Henry  8  apples  at  one  time,  6  at  another,  and  4  at  another ; 
how  many  did  he  give  him  ?     8  +  6+4  =  how  many  ?     Lent 
my  brother  10  dollars  at  one  time,  6  at  another,  and  4  at  another  ; 
how  many  dollars  did  I  lend  him?     10+6  +  4  =  how  many? 
Alfred  has  four  brothers  ;  to  one,  he  gave  1 1  pears,  to  another, 
9,  to  another,  6,  and  to  another,  3  ;  how  many  pears  did  he 
give  away  ?     11  +  9  +  6  +  3=  how  many  ?     He  also  gave 
his  two  sisters  5  each ;  how  many  did  he  give  to  both  brothers 
and  sisters  ?     11  +  9  +  6  +  3  +  5  +  5  =  how  many  ?     A 
man  bought  four  bushels  of  corn  for  3  dollars,  six  bushels  of 
oats  for  2  dollars,  and  9  bushels  of  wheat  for  12  ;  how  many 
bushels  of  grain  did  he  buy,  and  how  many  dollars  did  the 
whole  cost  him  ?     4  +  6  +  9  =  how  many  ?     3  +  2  +  12  = 
how  many  ?     John  has  9  dollars  ;  James,  12  ;  and  Joseph,  7  ; 
how  many  have  they  all?     9+12  +  7=  how  many?     Peter 


20  ADDITION  OF  SIMPLE  NUMBERS. 

gave  a  poor  man  11  cents  ;  Charles  gave  him  12  cents  ;  Hen- 
ry, 10;  and  George,  8;  how  many  did  they  all  give  him; 
ll  +  12  +  10-|-8=:  how  many?  A  man  was  three  days 
performing  a  journey ;  the  first  day  he  traveled  13  miles  ;  the 
second  day,  14;  and  the  third,  15;  how  far  did  he  travel? 
13  +  14+15  —  how  many  ?  How  many  sums  will  you  have 
done,  if  you  do  12  to-day,  15  to-morrow,  and  18  the  next  day? 
12+15+18=how  many? 

How  many  are  3  +  4?  13  +  4?  23  +  4?  33  +  4?  43+4? 
53  +  4?  63  +  4?  73+4?  83  +  4?  93  +  4?  How  many 
are4  +  4?  14  +  4?  24  +  4?  34  +  4?  &c.  How  many  are 
5  +  4?  15  +  4?  25  +  4?  35  +  4  ?  &c.  How  many  are  6  +  4  ? 
16  +  4?  26  +  4?  36  +  4  ?  &c.  How  many  are  7  +  4  ?  17+4? 
27+4?  37+4?  &c.  How  many  are  4 +  4?  4+14?  4+24? 
&c.  4+5?  4+15?  4  +  25?  &c.  4+6?  4  +  16?  4+26? 
&c.  4+7?  4+17?  4+27?  &c.  How  many  are  5  +  5  ? 

5  +  15?    5  +  25?  &c.     5  +  6?    5  +  16?    5+26?  &c.     6  +  6? 

6  +  16?  6+26  ?&c.     How  many  are  7  +  7?  7+17?  7+27? 
&c.     8+8?    8+18?     8+28?  &c.     9+9?     9+19?     9+29? 
&c.     Howmany  are  10+10?   10+20?  &c.   11+5?   11  +  15? 
12+4?   12  +  14?  &c. 

Questions  of  this  character  may  be  proposed  to  any  extent, 
and  should  in  no  instance  be  omitted  until  the  scholar  can  add 
without  hesitation.  Others  of  a  more  promiscuous  character 
should  likewise  be  proposed,  such  as  the,  following  : 

7  +  4  +  6=how  many?  2  + 1  +  9  +  6  +  7  +  6=  how  many? 
5  +  6  +  9  +  3=how  many?  6  +  4  +  2  +  8+1  +  3  +  5  +  7+9  = 
how  many?  11+9  +  6  +  12  =  how  many  ?  7+6  +  9  +  8  = 
how  many?  8  +  8  +  6  +  6  +  4  +  4=howmany?  7  +  7+9  +  9  = 
how  many  ?  2  +  4  +  6  +  8  +  10+1 2 =how  many  ? 

The  scholar  may  now  commence  the  use  of  the  slate  and 
pencil,  or  the  practice  of  written  arithmetic. 

The  first  consideration  to  which  his  attention  should  here 
be  directed,*is,  that  like  things  only  can  be  added  to  or  sub- 
tracted from  each  other.  It  would  be  absurd  to  attempt  to 
add  together  books  and  chairs,  to  see  how  many  books,  or  how 
many  chairs  the  whole  would  make  ;  the  number  of  each  would 
evidently  remain  unaffected.  If,  however,  we  add  books  to 
books,  we  obtain  a  number  greater  than  either  of  the  original 
numbers ;  that  is,  just  equal  to  them  both.  Neither  can  we 
add  units  to  tens ;  for  the  amount  would  be  neither  units  nor 
tens  ;  but  units  must  be  added  to  units,  tens  to  tens,  and  hun- 
dreds to  hundreds,  and  so  on.  But  ten  units  make  one  ten,  ten 


ADDITION  OF  SIMPLE  NUMBERS.  21 

tens  make  one  hundred,  and  ten  hundreds  make  one  thousand, 
&c. ;  that  is,  simple  numbers  increase  and  decrease  in  a  ten-fold 
ratio.  Hence  10  in  the  column  of  units  is  equal  to  1  in  the 
column  of  tens  ;  and  10  in  the  column  of  tens,  is  equal  to  1  in 
the  column  of  hundreds.  If,  then,  in  adding  up  the  column  of 
units  the  whole  should  amount  to  just  10,  it  is  obvious  that 
nothing  is  lost,  if  these  10  units  are  converted  into  1  ten,  and 
the  1  ten  added  to  the  column  of  tens ;  for  10  units  =  1  ten. 
The  same  is  true  of  other  denominations.  Therefore  the  fol- 
lowing General  Rule  will  be  found  applicable  to  Simple  Ad- 
dition : 

1st.  Write  down  the  numbers,  placing  units  under  units, 
tens  under  tens,  &c. 

2d.  Draw  a  line  underneath,  and  commence  at  the.  right 
hand  and  add  together  all  the  figures  in  the  first  column. 

3d.  If  the  sum  be  less  than  10,  set  it  down  at  the  foot  of  that 
column ;  if  it  be  10,  or  more  than  10,  it  will  consist  of  two 
figures  at  least ;  set  down  the  right  hand  one  as  before,  and  add 
the  left  hand  one  to  the  next  column,  it  being  in  all  cases  so 
many  tens,  when  compared  with  the  figures  added. 

4th.  Continue  to  perform  the  same  operation  with  the  re- 
maining columns,  observing  only  to  write  down  the  whole 
amount  of  the  left  hand  column. 

5th.  To  detect  any  error  that  may  have  been  committed, 
commence  again  and  add  each  column  downwards  ;  if  the 
same  numbers  are  obtained  by  each  operation,  the  work  is 
probably  right. 

Now  to  apply  this  rule,  let  us  add  together  the  four  following 
numbers,  viz.  1234  ;  2345  ;  6420  ;  and  5796.     The  rule  says, 
write  these  numbers  with  units  under  units,  tens 
under  tens,  &c.  thus  :  1234 

Now  to  add  these  numbers,  I  commence  at  the  ^2345 
right  hand,  and  first  add  together  the  unit  figures  ;  ^  6  4  2  0 
viz.  6,  0,  5,  4;  the  sum  I  find  to  be  15  units,  5796 

equal  to  1  ten  and  5  units ;  I  write  down  the 

5  units,  but  add  the  1  ten  to  the  next  column;  15795 
thus,  1  added  to  9  is  10,  and  2  are  12  and  4  are 
16  and  3  are  19 ;  as  before,  I  write  down  the  9,  and  add  the  1 
to  the  next  column  ;  this  being  added  gives  the  amount  17  ;  the 
7  is  written  down  and  the  1  again  carried  to  the  next  and  last 
column;  and  here  the  amount  is  15,  which  being  the  last 
column,  the  whole  number  is  written  down.  This  operation 


22  ADDITION  OF  SIMPLE  NUMBERS. 

gives  the  amount  of  the  four  numbers,  15795.  The  scholar 
will  readily  comprehend  the  nature  of  Simple  Addition  ;  viz. 
that  it  consists  in  uniting  two  or  more  numbers  of  the  same  de- 
nomination, so  as  to  find  their  amount. 

The    scholar  will   carefully  examine   the  following  sums, 
which  are  added,  to  see  if  he  obtains  the  same  result. 

2. 

847396  In  this  second   example,  .in   the  unit 

364869  column,  there  are  3  to  carry,  because 
482436  there  are  3  tens  or  30.  For  the  same 
622439  reason  there  are  2  to  carry  in  all  the  re- 
maining columns  but  one. 


2317140 


3  4. 

456789460  123456789 

680246802  987654321 

135791357  123456789 

423650825  987654321 

371574628  123456789 

856456333  987654321 


2924509405    3333333330 
5.  6.  7. 


2468024 
1357912 
4208642 
2135791 
3  3  5^5  7  7  9 
8866448 

9796365 
5636979 
8246024 
5963816 
4674364 
5796378 

1798670 
6124356 
7212345 
8463738 
5739168 
9156423 

22392596   40113926 

8.  9.  10. 

9176435  ^3457096  70819634 

5683214  61728349  64753278 

2345678  28394563  86542365 

0123456  56831234  23456789 


ADDITION  OF  SIMPLE  NUMBERS.  23 

11.  12.  13. 

81828384  39724638  45768903 

18283848  97236452  78924683 

99999999  15707934  13579246 

77777777  65972109  97576887 


14.  15.                                 16. 

123456789  63840596  567347205379 

987654321  27690903  275684467903 

546372819  46379108  123456789012 

951486804  55681396  378965842687 

468046872  63705418  593748265379 

608642109  36485074  916547681234 


17. 

18. 

19. 

795735198 

369248754 

1357924680 

49157038 

62958651 

261595827 

8930576 

2.5  8  1  7  3  1 

4230917546 

697005 

407347 

6083419 

29510 

80478 

682 

4812 

6184 

8472166264 

863 

236 

4462352 

2  7 

46 

44605 

4 

8 

2  1  0 

20.  A  farmer  sold  his  wheat  for  320  dollars ;  his  corn  for 
275  dollars  ;  his  oats  for  78  dollars  ;  his  barley  for  162  dollars  ; 
and  one  horse  for  132  dollars.     What  was  the  amount  of  his 
sales?     Ans.  $967. 

21.  A  merchant  owned   four  vessels,  which  were  worth, 
the  first,  $4800 ;  the   second,  $5200 ;  the  third,  $6000 ;  and 
the  fourth,  $6800  ;  he  had  also  goods  on  board  one  of  these 
vessels,  worth  $2700 ;  besides  $3500  deposited  in  the  bank. 
What  is  the  amount  of  his  property  ?     Ans.  $29000. 

22.  Three  farmers  have  each  562  acres  of  land  ;  how  many 
have  they  all?     Ans.  1686  acres. 

23.  A  farmer  fattened  and  killed  an  ox  for  market ;  the  hind 
quarters  weighed,  the  one  182  pounds,  and  the  other  177; 
each  of  the  fore  quarters  weighed  163  pounds ;  the  hide,  116 
pounds  ;  and  the  tallow,  120  pounds.     What  was  the  weight 
of  the  ox?     Ans.  921  pounds. 


24  ADDITION  OF  SIMPLE  NUMBERS. 

24.  In  a  certain  town  there  are  five  schools,  containing  the 
following  numbers  of  scholars,  viz.  72, 28,  65,  84,  and  91  ;  how 
many  children  are  attending  these  five  schools  1    Ans.  340.. 

25.  A  steamboat  performed  in  one  week  four  trips   from 
Hartford  to  New  York ;  on  the  first  trip,  she  took  for  passen- 
gers, $378,  for  freight,  $175  ;  on  the  second  trip,  for  passen- 
gers, $402,  for  freight,  $278  ;  on  the  third,  for  passengers, 
$263,  fpr  freight,  $147  ;  on  her  fourth  trip,  for  passengers  and 
freight,  $500.     What  did  her  bills  amount  to  for  the  week  ? 
Ans.  $2143. 

26.  A  carpenter  contracted  for  the  building  of  five  dwell- 
ings in  one  spring ;  for  the  first  he  was  to  receive  $1800 ;  for 
the  second,  $2100 ;  for  the  third,  $2221 ;  for  the  fourth,  $2850 ; 
and  for  the  fifth,  $3172.     To  what  did  his  contracts  amount? 
Ans.  $12143. 

27.  In  1834,  A  traveled  1320  miles;  in  1835,  he  traveled 
1162  miles;  in  1836,  2100  miles;  in  1837,  1400  miles;  in 
1838,  1992  miles.     How  many  miles  did  he  travel  from  1834 
to  1838  inclusive  ?     Ans.  7974. 

28.  Bought  of  my  neighbor   four  loads  of  hay;  the  first 
weighed  1600  Ibs. ;  the  second,  2100  Ibs. ;  the  third,  1999  Ibs. ; 
and  the  fourth,  1709  Ibs.  *  What  was  the  whole  weight  ?     Ans. 
7408  pounds. 

29.  A  wholesale  dealer  in  grain  has  in  one  bin  242  bushels 
of  wheat ;  in  another  he  has  2356  bushels  of  rye  ;  in  a  third, 
1556  bushels  of  oats  ;  in  a  fourth,  876  bushels  of  barley.    How 
many  bushels  of  grain  has  he  of  all  kinds  ?     Ans.  5030. 

30.  Suppose  I  am  indebted  to  A,  $2560 ;  to  B,  $27 ;  to  C, 
$169  ;  to  D,  $3470  ;  and  to  E,  $17.     How  much  do  I  owe  in 
all  ?     Ans.  $6243. 

31.  A  man  having  three  sons  and  two  daughters,  gave  to 
each  of  his  sons,  $379 ;  and  to  each  of  his  daughters,  $199. 
How  much  money  did  he  give  them  all  1     Ans.  $1535. 

32.  A  gentleman  being  asked  how  old  he  was,  said,  he 
was  married  when  he  was  29  years  of  age  ;  that  he  lived  with 
his  wife  8  years  before  the  birth  of  their  son,  who  was  now  27 
years  of  age.     What  was  the  father's  age  ?     Ans.  64  years. 

33.  A  man  bought  a  horse  for  $75  ;  a  chaise  for  $150  ;  and 
a  harness  for  $45  ;  he  then  sold  his  horse  for  $150 ;  his  chaise 
for  $125  ;  and  his  harness  for  $30.     What  did  he  pay  for  the 
whole,  and  what  did  he  receive  for  the  whole.     Ans.  Paid 
$270 ;  received  $305. 

34.  Add  togeth&t  three  hundred  and  seventy-five  thousand 


SIMPLE  SUBTRACTION. 


25 


and  sixty-five ;  nine  hundred  thousand  and  three ;  one  million  six 
hundred  thousand  seven  hundred  and  ninety-nine.  Ans.  2875867. 

35.  Add  also  ninety-nine  millions  ;  seven  hundred  and  fifty- 
five  millions  six  hundred  and  thirty-three.     Ans.  854000633. 

36.  There  are  5784  apples  in  one  pile ;  588  in  another ; 
84  in  a  third  ;  and  seven  hundred  and  seventy-nine  in  a  fourth. 
How  many  are  there  in  all  ?     Ans.  7235. 

Q.UESTIONS. — What  things  only  can  be  added  to  or  subtracted  from 
each  other  1  In  what  does  Simple  Addition  consist  1  Can  we  add 
units  to  tens  7  To  what  must  units,  &c.  be  added"?  For  what  num- 
ber do  we  carry  in  Simple  Addition  1  Why  for  10 1  Ans.  Because 
numbers  increase  in  a  ten-fold  ratio.  10  units  equal  how  many  10's  1 
What  is  the  Rule  for  Addition  7  How  do  you  write  down  the  num- 
bers 1  Where  do  you  commence  to  add  1  If  the  sum  of  the  figures 
added  be  less  than  10,  what  is  to  be  done  7  What  if  it  be  10,  or  more 
than  10  1  What  is  observed  respecting  the  last  or  left  hand  column  7 
How  may  errors  in  Addition  be  detected  7 


SIMPLE    SUBTRACTION. 

This  rule  is  directly  the  reverse  of  the  preceding ;  while 
we  are  there  taught  to  unite  several  numbers  into  one,  we  are 
here  taught  the  operation  by  which  one  number  is  taken  from 
another.  A  familiar  acquaintance  with  the  following  table 
should  be  the  first  object  of  the  scholar : 

SUBTRACTION  TABLE. 


1  —  1=0 

2  —  2  =  0 

3  —  3  =  0 

4  —  4  =  0 

2  —  1  =  1 

3  —  2  =  1 

4  —  3  =  1 

5  —  4  =  1 

3  —  1  =  2 

4  —  2  —  2 

5  —  3  =  2 

6  —  4  =  2 

4  —  1  =  3 

5  —  2  =  3 

6  —  3  =  3 

7  —  4  =  3 

5  —  1  =  4 

6  —  2  =  4 

7  —  3  =  4 

8  —  4  =  4 

6  —  1  =  5 

7  —  2  =  5 

8  —  3  =  5 

9  —  4  =  5 

7  —  1  -  6 

8  —  2  =  6 

9  —  3  =  6 

10  —  4=  6 

8  —  1  =  7 

9  —  2  =  7 

10  —  3=  7 

11—4=  7 

9—1  =  8 

10  —  2=  8 

11  —  3  =  8 

12  —  4=  8 

10  —  1  =  9 

11—2=  9 

12  —  3=  9 

13  —  4=  9 

11  —  1  =  10 

12  —  2  =  10 

13  —  3  =  10 

14  —  4  =  10 

12  —  1  =  11 

13  —  2  =  11 

14  —  3  =  11 

15  —  4  =  11 

13  —  1  =  12 

14  —  2  =  12 

15  —  3  =  12 

16  —  4  =  12 

26 


SIMPLE  SUBTRACTION. 


5  —  5  =  0 

6  —  6  =  0 

7  —  7=  0 

8  —  8=  0 

6  —  5  =  1 

7  —  6  =  1 

8  —  7=  1 

9  —  8  =  1 

7  —  5  =  2 

8  —  6  =  2 

9  —  7  =  2 

10  —  8=  2 

8C       Q 
—  O  —   O 

9  —  6=  3 

]0  —  7  =  3 

11—8=  3 

9  —  5  =  4 

10  —  6=  4 

11—7=  4 

12  —  8=  4 

10  —  5=  5 

11—6  =  5 

12  —  7=  5 

13  —  8=  5 

11—5=  6 

12  —  6=  6 

13  —  7=  6 

14  —  8=  6 

12  —  5=  7 

13  —  6=  7 

14  —  7=  7 

15  —  8=  7 

13  —  5=  8 

14  —  6=  8 

15  —  7=  8 

16  —  8=  8 

14  —  5=  9 

15  —  6=  9 

16  —  7=  9 

17  —  8=  9 

15  —  5  =  10 

16  —  6  =  10 

17  —  7  =  10 

18  —  8  =  10 

16  —  5  =  11 

17  —  6  =  11 

18  —  7  =  11 

19  —  8  =  11 

17  —  5  =  12 

18  —  6  =  12 

19  —  7=  12 

20  —  8  =  12 

9  —  9  =  0 

10  —  10=  0 

11  —  11=  0 

12  —  12=  0 

10  —  9=  1 

11  —  10=  1 

12  —  11=  1 

13—12=  1 

11—9  =  2 

12  —  10=  2 

13  —  11=  2 

14  —  12=  2 

12  —  9=  3 

13  —  10=  3 

14  —  11=  3 

15—12=  3 

13  —  9  =  4 

14  —  10=  4 

15  —  11=  4 

16  —  12=  4 

14  —  9=  5 

15  —  10=  5 

16  —  11=  5 

17  —  12=  5 

15  —  9=  6 

16  —  10=  6 

17—11=  6 

18  —  12=  6 

16  —  9=  7 

17  —  10=  7 

18  —  11=  7 

19  —  12=  7 

17  —  9=  8 

18  —  10=  8 

19  —  11=  8 

20  —  12=  8 

18  —  9=  9 

19  —  10=  9 

20—11=  9 

21  —  12=  9 

19  —  9  =  10 

20—10=10 

21  —  11  =  10 

22  —  12  =  10 

20  —  9  =  11 

21  —  10  =  11 

22  —  11  =  11 

23  —  12  =  11 

21—  9  =  12 

22  —  10  =  12 

23  —  11  =  12 

24  —  12  =  12 

MENTAL    EXERCISES   IN   SUBTRACTION. 

Charles  had  10  marbles,  and  gave  5  of  them  to  his  brother ; 
how  many  had  he  left  ?  He  then  gave  3  to  his  sister  ;  how 
many  did  there  then  remain  ?  A  boy  had  15  cents,  and  lost 
5  of  them  ;  how  many  had  he  left  ?  John  has  13  apples  ;  how 
many  will  he  have,  if  he  gives  away  5  ?  Peter  had  1 9  mar- 
bles, and  gave  away  9  of  them  ;  how  many  had  he  left  ?  A 
boy  purchased  two  toys,  for  the  one  he  gave  20  cts.  and  for  the 
other  12  cts;  what  was  the  difference  in  the  cost  of  them? 
Out  of  23  plumbs  a  boy  selected  12  ;  how  many  remained  ?  A 
boy 'threw  15  pennies  into  the  air,  and  caught  6  of  them  as 
they  fell ;  how  many  reached  the  ground  ?  Sold  one  cow  for 
19  dollars  and  another  for  12  ;  how  much  more  did  I  receive 
for  one  than  for  the  other  ?  Charles  has  19  marbles,  and  James 
has  1 1 ;  how  many  has  Charles  more  than  James  ?  Sold  one 


SIMPLE  SUBTRACTION.  27 

knife  for  21  cents  and  another  for  12  cents  ;  what  was  the  dif- 
ference in  price ?  11  taken  from  19,  how  many  will  remain? 
8  from  19  ?  10  from  19  ?  7  from  19  ?  11  from  19  ?  6  from 
19?  12  from  19?  5  from  19?  13  from  19?  4  from  19? 

14  from  19  ?     3  from  19  ?     If  9  be  taken  from  18,  how  many 
will  remain?     8  from  18?     10  from  18?     7  from  18?     11 
from  18?     6  from  18?     12  from  18?     5  from   18?     13  from 
18?     4  from  18?     14  from  18?     3  from  18?     15  from  18? 

2  from  18  ?     16  from  18  ?     1  from  18  ?     17  from  18  ?     If  10 
be  taken  from  20,  how  many  will  remain  ?     9  from  20  ?     11 
from  20  ?     8  from  20  ?     12  from  20  ?     7  from  20  ?     13  from 
20  ?     6  from  20  ?     14  from  20  ?     5  from  20  ?     15  from  20  ? 

4  from  20  ?     16  from  20  ?     3  from  20  ?     17  from  20  ?     2  from 

20  ?     18  from  20  ?     19  from  20  ?     11  from  21  ?     10  from  21  ? 
12  from  21  ?     9  from  21  ?     13  from  21  ?    8  from  21  ?    14  from 

21  ?     7  from  21  ?     15  from  21  ?     6  from  21  ?     16  from  21  ? 

5  from  21  ?     17  from  21  ?     4  from  21  ?     18  from  21  ?     3  from 

21  ?     19  from  21  ?     2  from  21  ?     11  from  22  ?     10  from  22  ? 

12  from  22?     9  from  22?     13  from  22?     8  from  22?     14 
from  22  ?     7  from  22  ?     15  from  22  ?     6  from  22  ?     16  from 

22  ?     5  from  22  ?     17  from  22  ?     4  from  22  ?     18  from  22  ? 

3  from  22  ?     19  from  22  ?     2  from  22  ?     12  from  23  ?     11  from 

23  ?     13  from  23  ?     10  from  23  ?     14  from  23  ?     9  from  23  ? 

15  from  23?     8  from  23  ?     16  from  23?    7  from  23?     17  from 
23  ?     6  from  23  ?     18  from  23  ?     5  from  23  ?     19  from  23  ? 

4  from  23  ?     20  from  23  ?     3  from  23  ?  "21  from  23  ?     2  from 

23  ?     22  from  23  ?     1  from  23  ?     12  from  24  ?     11  from  24  ? 

13  from  24?     10  from  24?     14  from  24?     9  from  24?     15 
from  24?    8  from  24?     16  from  24?     7  from  24?     17  from  24? 

6  from  24  ?     18  from  24  ?    5  front  24  ?     19  from  24  ?     4  from 

24  ?     20  from  24  ?     3  from  24  ?     21  from  24  ?     2  from  24  ? 
22  from  24  ?     1  from  24  ?     23  from  24  ? 

16  boys  went  on  a  sailing  excursion,  only  7  of  them  return- 
ed ;  the  others  were  drowned ;  how  many  were  lost  ?  If 
from  a  pile  of  20  apples,  I  take  away  13,  how  many  will  be 
left  ?  A  man  started  on  a  journey  of  23  miles  ;  after  he  had 
traveled  16  miles,  he  stopped  to  feed  his  horse  ;  how  far  had 
he  then  to  travel?  A  man  having  22  chickens,  killed  13  of 
them;  how  many  were  left  ?  From  a  stick  of  timber  19  feet 
long,  7  feet  were  cut  off ;  what  was  the  length  of  the  remain- 
der ?  19  —  7=howmany?  21 — 9=howmany?  17  —  9 
r=how  many  ?  23  —  5 =how  many  ?  15  —  7  =  how  many  ? 
24  —  9=howmany?  11 — 7=howmany?  22  —  16=how 
many?  12— -.7= how  many  ?  13  —  5= how  many?  14  — 


28  SIMPLE  SUBTRACTION. 

7=howmany?  15  —  9=howinany?  16  —  9=howmany? 
17  —  8=howmany?  18  —  7z=howmany?  19  —  ll=:how 
many  ?  20  —  9 = how  many  ?  21  —  7 =how  many  ?  22 — 9 
r=how  many  ?  23  —  13=how  many  ? 

From  the  preceding  table  and  examples,  the  scholar  will  com- 
prehend the  nature  of  Sample  Subtraction  ;  his  next  step  will  be 
to  practice  with  his  slate  and  pencil.  It  will  already  ha.ve  been 
observed,  that  only  two  numbers  are  employed  in  a  single  ope- 
ration of  subtraction.  The  larger  of  these,  two  numbers  is  call- 
ed the  minuend  ;  and  the  smaller,  the  subtrahend.  The  object 
of  the  rule  is  to  find  the  difference  between  the  two  ;  that  is,  to 
find  how  much  will  remain  of  the  larger  after  the  smaller  is  taken 
from  it.  The  number  obtained  by  the  operation,  is  called  the 
remainder. 

The  scholar  may  be  guided  by  the  following  rule  : 

RULE. — 1st.  Write  the  less  of  the  two  numbers  under  the 
greater,  with  units  under  units  and  tens  under  tens,  dj-c.  and 
draw  a  line  beneath  them. 

2d.  Commence  with  the  right  hand  figure  of  the  lower  line  or 
subtrahend,  and  take  it  from  the  figure  which  stands  directly 
above  it,  if  practicable.  Do  the  same  with  the  remaining  fig- 
ures in  the  subtrahend,  if  practicable,  and  the  operation  will  be 
completed. 

3d.  But  whenever  this  cannot  be  done,  that  is,  when  the  low- 
er figure  is  the  larger,  10  should  be  added  to  the  upper  figure 
and  the  lower  one  taken  from  the  sum. 

4th.  Whenever  10  is  added  to  an  upper  figure,  I  must  be  car- 
ried or  added  to  the  next  lower  figure ;  that  is,  I  is  to  be  car- 
ried  whenever  10  is  borrowed. 

5th.  To  prove  the  work,  add  the  remainder  to  the  subtrahend, 
and  if  the  work  be  right,  the  amount  will  correspond  with  the 
minuend. 

The  scholar  will  easily  comprehend  the  nature  of  this  rule, 
unless  he  should  find  difficulty  in  understanding  why,  when  we 
borrow  10,  we  are  required  to  carry  only  1.  He  must  howev- 
er remember,  that  by  the  addition  of  this  10  to  the  upper  num- 
ber, he  has  increased  the  value  of  that  number  10  units,  10 
tens,  10  hundreds,  or  10  thousands,  according  to  the  place 
the  figure  occupies.  If  he  add  it  to  the  units,  the  value  of  the 
addition  is  1  ten ;  if  to  the  tens,  the  value  is  one  hundred,  be- 
cause 10  units  make  one  ten,  and  10  tens,  one  hundred,  &c. 
Now  by  the  rule,  if  10  be  borrowed,  1  must  be  carried  to  the 


SIMPLE  SUBTRACTION.  29 

next  lower  figure  ;  by  which  operation,  one  more  will  be  taken 
from  the  figure  in  the  minuend ;  and  this  one  more,  which  is 
thus  removed,  is  just  equal  in  value  to  the  10  that  was  added, 
for  it  is  taken  from  a  figure  one  degree  farther  to  the  left.  But 
this  subject  will  be  more 'clearly  comprehended,  when  illustra- 
ted by  example.  Take  the  following  sum  : 

In  this  example,  it  is  evident  that  if  From  635428 
6  be  taken  from  8,  2  will  remain  ;  and  Take  382516 
if  the  1  ten  be  taken  from  2  tens,  1 

ten  will  remain.  But  how  is  5  in  the  Rem.  252912 
place  of  hundreds  to  be  taken  from 

the  4  above  it  ?  Evidently  by  the  third  section  of  the  rule  ;  that 
is,  10  is  added  to  the  4  in  the  minuend,  by  which  addition,  it 
will  become  14  hundred,  from  which  if  5  hundred  be  taken,  9 
hundred  will  remain,  which  is  the  third  figure  in  the  remainder. 
But  by  this  operation  the  minuend  has  been  increased  10  hun- 
dred ;  if  therefore  I  add  1  to  the  2  thousand,  it  will  become  3 
thousand,  and  consequently,  when  subtracted  from  the  figure  5 
above  it,  will  take  one  thousand  more  from  the  minuend,  so  that 
only  2  will  remain.  If  therefore  10  hundred  was  in  one  in- 
stance added  to  the  minuend,  in  the  other,  1  thousand,  its  equal, 
has  been  taken  from  it.  The  same  reasoning-is  applicable  to  the 
8  ;  10  is  added  to  the  3,  which  increases  it  to  13  ;  the  8  is  ta- 
ken from  the  13j  and  5  remains.  There  is  then  one  to  carry  to 
the  3,  which  thus  increased  is  taken  from  the  6,  and  2  re- 
mains. The  whole  remainder  therefore  is  252912. 

Now  if  this  remainder  be  added  to  the  lower  number  or  sub- 
trahend, the  amount  will  be  the  minuend ;  which  proves  that 
the  operation  is  correct,  thus  : 

Subtrahend  =  382516 
Remainder  =252912 


Minuend      =635428 
2.  3.  4. 

From  66683     From  894673     From  987654321 
Take  25966     Take  768596      Take  123456789 


40717  126077  864197532 

5.  6. 

Min.  1000000000   Min.  1075608756 
Subtr.  999999999   Subtr.  698453874 


30  SIMPLE  SUBTRACTION. 

7.  8. 

Min.     9634657942          Min.     30000000 
Subtr.  8326547286  Subtr.  29999999 


9.  10. 

Min.     100000000         Min.    90807060504 
Subtr.  9         Subtr.  80906070400 


11.  12. 

Min.  965843125      56789357913 
Subtr.  428642086      4135724,3648 


13  14 

812345678946       7539753168 
488765432109        3640854279 


15.  16. 

23456789012   5678535347963 
92567845023   4756246125787 


17.  18. 

25386744466       66668888444 
18192939495       59999999999 


19.  20. 

63456956429         9888888888 
58686721398         8999999999 


21.  22. 

1111111111         6666666666 

8888888888         5777777777 


APPLICATION. 


23.  A  was  bora  in  1679  ;  how  old  was  lie  in  1777  ?     Ans. 
98  years. 

24.  From  1600000  take   900000,  and  from  the  remainder 
take  699999,  and  how  much  will  remain?     Ans.  1. 


SIMPLE  SUBTRACTION.  31 

25.  A  man  has  two  flocks  of  sheep ;  in  the  one  there  are  693, 
and  in  the  other,  499  ;    what  is  the  difference  in  these  flocks  1 
Ans.  194. 

26.  A  man  has  in  his  possession  property  to  the  amount  of 
$15728,  and  he  owes  $7869  ;  how  much  will  remain  in  his 
hands,  when  his  debts  are  paid  ?     Ans.  $7859. 

27.  America  was  discovered  in   1492.     How  long  will  it 
have  been  discovered  in  1846  ?     Ans.  354  years. 

28.  A  man  being  asked  how  old  he  was  when  his  eldest  son 
was  born,  said  that  his  own  age  was  79  years,  and  his  son's 
42  years  ;  what  was  his  age  at  the  birth  of  his  son  ?     Ans.  37 
years. 

29.  The  amount  of  A's  debts  was  2356  dollars  ;  the  amount 
of  his  property,  5672  dollars.     How  much  had  he  left  after  his 
debts  were  paid  ?     Ans.  $3316. 

30.  A  merchant  bought  a  quantity  of  cloth  for  $572,  and 
sold  it  for  $526.     Did  he  gain  or  lose,  and  how  much  ?     Ans. 
Lost  $46. 

31.  To  what  number  must  I  add  576  to  make  the   amount 
1726?      Ans.  1150. 

32.  Bought  cotton  in  one  month  to  the  value  of  $572896  ; 
and  sold  the  same  for  $600027.     How  much  did  I  gain  ?     Ans. 
$27131. 

33.  If  the  sum  of  two  numbers  be  2793,   and  one  of  those 
numbers,  ]892,  what  is  the  other?     Ans.  901. 

34.  A  merchant  bought  742  yards  of  cloth  and  sold  all  but 
7  yards.     How  much  did  he  sell  ?     Ans.  735  yards. 

35.  A  man  paid  1 182  dollars  for  a  house,  and  sold  the  same 
for  1069  dollars.     How  much  did  he  lose  ?      Ans.  $113. 

36.  A  farmer  purchased  a  farm,  for  which,  including  the 
buildings,  he  paid  $6782;  the  buildings  were  worth  $2896  ; 
what  was  the  value  of  the  land  ?     Ans.  $3886. 

37.  A  person  owed  a  merchant  $999,  and  paid  him  all  but 
$179.     How  much  did  he  pay  him  ?     Ans.  $820. 

Sums  requiring  in  their  solution,  the  application  of  both  Addi- 
tion and  Subtraction. 

38.  I  hold  in  my  possession  a  note  for  $560,  on  which  there 
is  due  $70  interest.     On  the  back  are  two  endorsements  ;  one, 
$320,  and  the  other,  $260.     What  is  now  due  ?     Ans.  $50. 

39.  There  are  $  1000  in  four  different  purses ;  in  the  first  there 


32  SIMPLE  SUBTRACTION. 

are  $96  ;  in  the  second,  $310  ;  in  the  third,  $205.     How  many 
are  there  in  the  fourth  ?     Ans.  $389. 

40.  Four  men  agreed  to  contribute  for  a  benevolent  object, 
as  follows  ;  the  first,  $34  ;  the  second,  $50  ;  the  third,  $100 ; 
and  the  fourth,  $150.     Three   of  them  having  paid,  the  sum 
amounted  to  $234 ;  which  subscription  was  unpaid  ?     Ans.  The 
third. 

41.  A  had  172  yards  of  cloth,  of  which  he   sold  57  to  B, 
and  42  to  C.     How  many  yards  were  left  ?     Ans.  73. 

42.  A  man  having  $3986,  paid  one  debt  of  $1997,  and  another 
of  $1089.     How  many  dollars  had  he  left  ?     Ans.  $900. 

43.  A  man  at  death  left  an  estate  of  $9876.     In  his  will  he 
gave  to  each  of  his  three  sons,  $1800  ;  to  his  daughter,  $1500  ; 
and  the  remaining  part  he  left  to  his  wife.     What  was  the 
wife's  portion  ?     Ans.  $2976. 

44.  A  person    commencing   business    found  that  he  had 
$790  in  money ;  in  goods,  $1260  ;   he  held  also  three  notes  of 
$150  each.     After  trading  six  years,  he  retired  from  business, 
and  found  that  his  property  amounted  to  $6000.     How  much 
had  he  gained  by  trading?     Ans.  $3500. 

45.  Bought  four  chests  of  tea,  weighing  72,  79,  83,  and  87 
pounds.    From  these  I  sold,  to  one  man,  46,  to  another,  95,  and 
to   a  third,    113  pounds.     How  much  tea  had  I  remaining? 
Ans.  67  pounds. 

46.  A  man  owed  for  his  farm,  $2100  ;   for  house  furniture, 
$156  ;  for  a  horse,  $96  ;  for  a  yoke  of  oxen,  $120  ;  for  a  flock 
of  sheep,  $86.     In  one  year  he  sold  from  his  farm  grain  to  the 
value  of  $462  ;  butter  and  cheese  to  the  value  of  $156  ;    stock 
to  the  amount  of  $320.     How  much  did  he  owe  at  the  end  of 
the  year?     Ans.  $1620. 

47.  A  man  received  $7000  as  a  legacy  ;  he  was  previously 
worth  $8560;  he  then  commenced  traveling,  and  in  7  years 
he  spent    $9873.     How   much   was   he  then  worth?     Ans. 
$5687. 

48.  A  man  being  asked  how  old  he  was,  replied,  that  he 
married  at  21  years  of  age,  and  that  in  19  years  more  he  should 
have  been  married  60  years.     How  old  was  he  ?    Ans.  62 
years. 

49.  Bought  1000  pounds  of  coffee  ;  from  this  quantity  I  sold 
at  one  time,  376  pounds  ;    and  at  another,  512  pounds.     How 
much  had  I  remaining?     Ans.  112  pounds. 

50.  A  man  bought  two  hogsheads  of  melasses,  the  one  con- 


SIMPLE    MULTIPLICATION. 


33 


taining  65  gallons,  and  the  other  69  gallons  ;    from  the  two  he 
sold  112  gallons.     How  much  had  he  left?     Ans.  22  gallons. 

QUESTIONS. — How  does  this  rule  compare  with  Addition  1  What 
are  we  taught  in  Subtraction  ?  How  many  numbers  are  employed  in 
this  rule?  What  are  they  called?  What  is  the  object  of  the  rule? 
What  name  is  given  to  what  is  left  after  the  operation  1  What  is  the 
rule  for  subtraction  1  What  is  to  be  done  when  the  lower  figure  is  the 
larger?  When  is  1  to  be  carried?  Do  you  ever  have  more  than  one 
to  carry  in  subtraction?  Ans.  We  do  not.  How  do  you  prove  the 
work  ?  How  will  you  show  that  carrying  one  as  directed,  is  equiva- 
lent to  the  ten  borrowed  ? 


SIMPLE    MULTIPLICATION. 

The  rule  to  which  the  scholar's  attention  will  now  be  direct- 
ed, is  one  by  which  a  number  is  produced  from  two  given  num- 
bers, which  shall  contain  either  of  these  given  numbers  as  many 
times  as  there  are  units  in  the  other ;  or,  it  is  the  repeating  of 
one  number  as  many  times  as  there  are  units  in  the  other.  For 
example,  let  8  and  4  be  the  numbers  ;  that  is,  let  8  be  repeated 
4  times.  The  result  of  these  four  repetitions  of  8  is  obviously 
32.  But  32  contains  8  four  times,  or  4  eight  times  ;  or,  in  other 
words,  32  contains  either  number  as  many  times  as  there  are 
units  in  the  other.  The  scholar  must  first  learn  the  follow- 
ing table : 

MULTIPLICATION  TABLE. 


1X1=1 

2X1=2 

3x1=3 

4x    1=    4 

1X2=2 

2x2=4 

3x2=6 

4x    2=    8 

1X3=3 

2x3=6 

3x3=9 

4x    3  =  12 

1X4=4 

2x4=8 

3x    4  =  12 

4x    4  =  16 

1X5=5 

2  X    5  =  10 

3  X    5  =  15 

4  X    5  =  20 

1X6=6 

2x    6  =  12 

3x    6  =  18 

4x    6  =  24 

1X7=7 

2x    7=14 

3X    7  =  21 

4x    7  =  28 

1X8=8 

2x    8  =  16 

3x    8  =  24 

4x    8  =  32 

1X9=9 

2x    9  =  18 

3x    9  =  27 

4x    9  =  36 

1  x  10=10 

2  X  10  =  20 

3  X  10  =  30 

4  x  10  =  40 

1  X  11  =  11 

2  Xll=22 

3  xll=33 

4  X  11  =44 

1  X  12  =  12 

2  Xl2=24 

3x12  =  36 

4  X  12  =  48 

34 


SIMPLE  MULTIPLICATION. 


5X    1=    5 

6x    1=    6 

7X1=7 

8x    1=    8 

5x    2  =  10 

6x    2  =  12 

7x    2  =  14 

8x    2=16 

5  X    3  =  15 

6x    3  =  18 

7x    3=21 

8x    3  =  24 

5x    4  =  20 

6  X    4  =  24 

7x    4=28 

8  x    4  =  32 

5x    5  =  25 

6  X    5  =  30 

7x    5=35 

8x    5  =  40 

5x    6  =  30 

6  x    6  =  36 

7  X    6  =  42 

8x    6  =  48 

5x    7  =  35 

6  x    7  =  42 

7x    7=49 

8x    7  =  56 

5X    8  =  40 

6  X    8  =  48 

7x    8  =  56 

8x    8  =  64 

5x    9  =  45 

6  x    9  =  54 

7x    9=63 

8  x    9  =  72 

5  X  10  =  50 

6  x  10  =  60 

7  X  10  =  70 

8x10  =  80 

5  X  11=55 

6  X  11=66 

7x  11=77 

8  x  11  =  88 

5  X  12  =  60 

6  X  12  =  72 

7  x  12  =  84 

8  X  12  =  96 

9X    1=     9 

10x    1=   10 

11X    1=   11 

12  X    1=   12 

9X    2=   18 

10X   2=  20 

11  X   2=  22 

12  X   2=  24 

9X    3=  27 

10x    3=   30 

11  X    3=   33 

12X    3=   36 

9X   4=  36 

10X    4=  40 

11  X   4=   44 

12  x    4=  48 

9X   5=  45 

10x    5  =  50 

11  X    5=   55 

12  X    5=   60 

9x    6=   54 

10x    6=   60 

11  X    6=   66 

12  X    6=   72 

9X   7=  63 

10  X    7=  70 

11  X    7=   77 

12  X    7=   84 

9x   8=  72 

10  X    8=   80 

11  X    8=   88 

12  X    8=   96 

9x    9=   81 

10  X    9=  90 

11  X    9=   99 

12X    9  =  108 

9x10=   90 

10x10=100 

11X10=110 

12x10  =  120 

9x11=   99 

10x11  =  110 

11X11  =  121 

12x11=132 

9x12  =  108 

10x12  =  120 

11X12  =  132 

12x12  =  144 

MENTAL  EXERCISES  IN  MULTIPLICATION. 

Bought  2  apples  for  3  cents  apiece  ;  what  did  they  cost  ? 
2  x  3=how  many  ?  Gave  to  each  of  3  boys,  4  plums  ;  how 
many  did  I  give  away?  3x4=how  many?  What  cost  4 
oranges  at  5  cents  apiece  ?  4  x  5=how  many  ?  What  cost  6 
oranges  at  5  cents  apiece  ?  6  x  5=how  many  ?  What  cost  7 
quarts  of  cherries  at  6  cents  a  quart?  7x6=how  many?  8 
quarts  of  melasses  at  7  cents  per  quart?  8x7=how  many? 

9  picture  books  at  8  cents  apiece  ?    9  x  8=how  many  ?     What 
cost  10  pints  of  wine  at  9  cents  a  pint  ?    10  x  9= how  many  ? 

11  yards  of  muslin  at  10  cents  a  yard  ?    11  X  10=how  many  ? 

12  yards  at  11  cents  a  yard?     What  cost  12  loads  of  straw  at 
5  dollars  a  load  ?    12  X  5=  how  many  ?    at  9  dollars  a  load  ?    7 
dollars?    6  dollars?    8  dollars?    10  dollars?    12  dollars?    11 
dollars  ?    What  will  it  cost  to  ride  6  miles  at  5   cents  a  mile  ? 
What  to  ride  7  miles  ?    9  miles  ?    4  miles  ?    5  miles  ?    8  miles  ? 

10  miles  ?    1 1  miles  ?    12  miles  ?     Six  men  paid  a  poor  man  6 


SIMPLE  MULTIPLICATION.  35 

dollars  each ;  how  much  did  the  poor  man  receive  ?  How 
much  would  he  have  received  had  they  paid  9  dollars  each  ? 
7  dollars  each ?  5  dollars  each?  10  dollars  each?  12  dollars 
each  ?  7  men  purchased  a  horse  in  company  and  paid  9  dol- 
lars each  ;  what  did  they  pay  for  the  horse  ?  What  would 
the  horse  have  cost  them,  had  they  paid  only  7  dollars  each  ?  If 
they  had  paid  8  each?  11?  10?  12  ?  6  ?  If  a  man  travel  5  miles 
in  1  hour,  how  far  will  he  travel  in  7  hours  ?  in  5  hours  ?  in 
9  hours  ?  in  6  hours  ?  in  8  hours  ?  in  10  hours  ?  in  1 1  hours  ? 
What  is  the  product  of  6  multiplied  by  1 1  ?  by  5  ?  by  9  ?  by 
7?  by  8?  by  6?  by  10?  by  11  ?  by  12  ?  Whatisthe  pro- 
duct of  9  multiplied  by  3  ?  by  12  ?  by  4  ?  by  1 1  ?  by  5  ?  by 
10?  by  6?  by  9?  by  7?  by  8  ?  Of  12  multiplied  by  7  ?  by 
2?  by  12?  by  3?  by  11?  by  4  ?  by  10?  by  5  ?  by  9  ?  by 
6  ?  by  8  ?  Of  11  multiplied  by  11  ?  by  7  ?  by  8  ?  by  6  ?  by 
5?  by  9?  by  10?  by  12? 

If  a  man  earn  9  dollars  a  month,  what  will  he  earn  in  8 
months  ?  If  the  board  of  a  man  and  his  wife  be  7  dollars  per 
week,  what  will  it  amount  to  in  1 1  weeks  ?  If  one  load  of  hay 
cost  11  dollars,  what  will  12  loads  cost?  If  a  yard  of  ribbon 
cost  8  cents,  what  will  1 1  yards  cost  ? 

It  has  already  been  said  that  two  numbers  are  required  for  the 
operation,  viz.  the  multiplicand,  or  number  to  be  multiplied  or 
repeated ;  and  the  multiplier,  or  number  showing  how  many 
times  the  multiplicand  is  to  be  taken  or  repeated.  The  multipli- 
er and  multiplicand,  when  spoken  of  together,  are  called  factors. 
The  number  obtained  by  the  operation  is  called  the  product. 
A  short  illustration  will  show  this  rule  to  be  an  abbreviation 
of  addition.  Suppose  it  be  required  to  multiply  5  by  4.  If  the 
scholar  turn  to  his  table  he  will  find  the  product  of  these  two 
numbers  is  20.  The  same  result  is  obtained  if  four  5's  be 
added  together;  thus,  5  +  5  +  5  +  5  =  20.  It  will  then  be  per- 
ceived that  if  one  of  the  two  numbers  which  are  to  be  mul- 
tiplied together,  be  written  down  as  many  times  as  there  are 
units  in  the  other,  and  these  several  numbers  be  then  added,  we 
obtain  the  same  result  as  when  these  two  numbers  are  multi- 
plied together.  Multiplication  is  therefore  a  short  method  of 
performing  addition. 

It  is  highly  important  that  the  scholar  should  obtain  accurate 
views  of  the  value  of  the  product  arising  from  the  multiplication 
of  any  two  numbers.  There  will  be  no  difficulty  in  this  respect, 
when  it  is  required  to  multiply  unit  figures  only,  or  when  a 


36  SIMPLE  MULTIPLICATION. 

unit  figure  only  is  given  as  a  multiplier ;  for  then  the  product 
of  each  figure  will  be  of  the  same  denomination  as  the  figure  it- 
self. But  when  the  two  factors  consist  each  of  several  figures, 
so  that  tens  are  to  be  multiplied  by  tens,  and  hundreds  by  hun- 
dreds, the  scholar  will  not  so  readily  comprehend  the  nature 
of  the  operation.  He  must  however  remember,  that  when  his 
multiplying  figure  is  tens,  it  will  raise  the  value  of  the  product 
of  each  figure  in  the  multiplicand  one  degree ;  when  it  is 
hundreds,  it  will  raise  the  value  of  each  two  degrees  ;  and  when 
thousands,  three  degrees,  &c.  Let  the  scholar  carefully  notice 
what  is  here  statej.  If  tens  in  the  multiplier  be  multiplied 
into  units  in  the  multiplicand,  the  product  is  tens  ;  if  into  tens, 
the  product  is  hundreds.  If  hundreds  in  the  multiplier  be  mul- 
tiplied into  units  in  the  multiplicand,  the  product  is  hundreds  ; 
if  into  tens,  the  product  is  thousands  ;  and  if  into  hundreds,  the 
product  is  tens  of  thousands,  &c.  This  explanation  will  en- 
able the  scholar  to  understand  the  rule  under  Case  3d,  relative 
to  writing  down  the  several  products.  He  will  readily  perceive 
it  to  be  nothing  more  or  less  than  writing  units  under  units, 
and  tens  under  tens,  &c. 

CASE  1st. — WHEN  THE  MULTIPLIER  DOES  NOT  EXCEED  12. 

RULE. — Commence  at  the  right  hand  and  multiply  each  fig- 
ure in  the  multiplicand  by  the  multiplier,  carrying  and  setting 
down  as  in  the  preceding  rules. 

Ex.  1.  Multiply  452  by  3. 

OPERATION. 
452 

3 

Prod.  1356  I  first  say,  3  times  2  are  6,  which,  being 
less  than  10,  I  set  down;  next,  3  times  5  are  15,  which  is  5 
units  and  1  ten  ;  the  units  I  set  down  and  carry  the  ten ;  thus, 
3  times  4  are  12  and  one  to  carry  are  13.  I  thus  find  the  whole 
product  to  be  1356.  This  number  must  consequently  contain 
the  multiplicand  3  times,  and  may  be  obtained  by  adding  to- 
gether three  452's  ;  thus,  4  5  2 
The  same  result  is  therefore  obtained  by  multiplica-  452 
tion  as  by  addition,  but  more  expeditiously.  The  452 
operation  is  proved  by  dividing  the  product  by  the 
multiplier,  which,  if  the  work  be  correct,  will  give  1356 
the  multiplicand.  The  scholar  is  not,  however,  supposed  yet  to 


SIMPLE    MULTIPLICATION.  37 

understand   division,  and  will   not   be  required   to   prove    his 
work,  till  more  advanced. 

2.  Multiply  6432  by  4.  OPERATION. 

6432 
4 


2572  8  =  Prod. 


3.  Multiply  123456  by  6.  OPERATION. 

123456 
6 


74073  6  =  Prod. 


4.  Multiply  234567  by  8.  OPERATION. 

234567 

8 


187653  6  =  Prod. 

5.  Multiply  345678  by  10.     Prod.  3456780. 

6.  Multiply  456789  by  12.     Prod.  5481468. 

7.  Multiply  729468  by  5.     Prod.  3647340. 

8.  Multiply  295105538  by  7.     Prod.  2065738766. 

9.  Multiply  4285637  by  9.     Prod.  38570733. 

10.  Multiply  462838  by  11.     Prod.  5091218. 

11.  Multiply  99887766  by  9.     Prod.  898989894. 

12.  Multiply  765987879  by  7.     Prod.  5361915153. 

13.  Multiply  9864579  by  8.     Prod.  78916632. 

14.  Multiply  7799886655  by  12.     Prod.  93598639860. 

APPLICATION. 

15.  Bought  72  yards  of  cloth  for  three  dollars  per  yard. 
What  did  it  cost?     Ans.  $216. 

16.  Sold  137   sheep  at  5  dollars  per  head.     How  much  did 
I  receive  ?     Ans.  $685. 

17.  Employed  196  men  one  week,  for  $8  per  week.     How 
much  did  I  pay  them  all?     Ans.  $1568. 

18.  How  many  miles  will  a  man  travel  in  297  days,  if  he 
travel  12  miles  a  day  ?     Ans.  3564. 

19.  If  I  take  11  steps  in  one  minute,  how  many  steps  shall  I 
take  in  2  hours  56  minutes,  or  176  minutes  ?     Ans.  1 936  steps. 

20.  If  a  horse  trot  7  miles  in  one  hour,  how  far  will  he  trot 
in  76  hours  ?     Ans.  532  miles. 

4 


38  SIMPLE    MULTIPLICATION. 

CASE  3d. — WHEN  THE  MULTIPLIER  is  A  COMPOSITE  NUMBER. 

NOTE. — A  composite  number  is  one  which  can  be  produced 
by  the  multiplication  of  two  or  more  small  numbers ;  thus,  6  and 
3  are  the  component  parts  of  18  ;  because  6x3  =  18. 

RULE. — Multiply  first  by  one  of  the  component  parts  of  the 
multiplier,  and  this  product  by  the  other  component  part ;  the  last 
product  will  be  the  one  sought. 

Ex.  1.  Multiply  4568  by  24.  The  component  parts  of  24 
are  6  and  4,  therefore, 

4568 
6 


2  7  4  0  8= six  times  the  multiplicand. 
4 


10963  2 =24  times  the  multiplicand. 
Ex.  2.  Multiply  459684  by  36.     36=4  x  9,  therefore, 

459684 
9 

4137156=9  times  the  multiplicand. 
4 


1654862  4  =  36  times  the  multiplicand. 

It  is  immaterial  in  what  order  the  component  parts  are  taken. 

3.  Multiply  5634286  by  18.  Prod.  101417148. 

4.  Multiply  4327648  by  27.  Prod.  116846496. 

5.  Multiply  7295678  by  36.  Prod.  262644408. 

6.  Multiply  4639546  by  48.  Prod.  222698208. 

7.  Multiply  3695475  by  42.  Prod.  155209950. 

8.  Multiply  54639578  by  60.  Prod.  3278374680. 

9.  Multiply  578016937  by  96.  Prod.  55489625952. 
10.  Multiply  79875643  by  63.  Prod.  5032165509. 

APPLICATION. 

11.  Bought  75  tons  of  hay  at  $15  per  ton.     What  did  the 
whole  cost?    Ans.  $1125. 

12.  How  many  hours  are  there  in  76  days  ?     Ans.  1824. 

13.  How  many  minutes  are  there  in  49  hours  ?     Ans.  2940. 


SIMPLE    MULTIPLICATION.  39 

14.  How  many  days  are  there  in  21  years  ?     Ans.  7665. 

15.  What  cost  172  acres  of  land  at  $36  per  acre  ?  Ans.  6192. 

16.  What  will  876  pounds  of  coffee  cost  at  28  cents  per 
pound.     Ans.  24528  cts. 

CASE   3d. — WHEN  THE  MULTIPLIER  EXCEEDS  12,  AND  is  NOT 

A  COMPOSITE    NUMBER. 

RULE. — Multiply  each  figure  in  the  multiplicand  by  each 
figure  in  the  multiplier  separately,  commencing  with  the  right 
hand  figure  of  each,  and  set  down  the  first  figure  of  each  product 
directly  under  the  multiplying  figure.  After  each  figure  in  the 
multiplier  has  been  taken,  add  together  the  several  products ; 
the  amount  will  be  the  required  product. 

Ex.  1.  342635  by  125. 

OPERATION. 

342635 
1   2  5 


171317  5  =  Product  of  5  units. 

68527  0  =  Product  of  2  tens  removed  1  place  to  the  left. 
342635  =  Product  of  1  hundred,  two  places  to  the  left. 

4282937  5=:Product  of  125. 
Ex.  2.  Multiply  167498  by  231. 

OPERATION. 

167498 
2  3  1 

16749  8  =  Product  of  1  unit. 

50249  4  =  Prod,  of  3  tens  removed  one  place  to  the  left. 
33499  6  =  Prod,  of  2  hundreds,  two  places  to  the  left. 

3  8  692  0~3~8  =  Prod.of  231. 

3.  Multiply  36598674  by  432.        PERFORMED. 

36598674 
432 


73197348 
109796022 
146394696 

1581062716  8=Prod. 


40  SIMPLE    MULTIPLICATION. 

4.  Multiply  46354897816  by  56843.     Prod.  26349514565- 
54888. 

5.  Multiply  3781 99886432  by  42395.     Prod.  1603378418- 
5284640. 

6.  Multiply  85698436946  by  46743.     Prod.  400580203816- 
6878. 

7.  Multiply  6739542  by  346.     Prod.  2331881532. 

8.  Multiply  72926495  by  4567.     Prod.  333055302665. 

9.  Multiply  89764267  by  999.     Prod.  89674502733. 

10.  Multiply  46371674  by  49684.     Prod.  2303930251016. 

11.  Multiply  8429638  by  7294.     Prod.  61485779572. 

12.  Multiply  7364951  by  888.     Prod.  6540076488. 

13.  There  are  69  pieces  of  cloth  containing  each  1 12  yards  ; 
how  many  yards  are  there  in  all  ?     Ans.  7728. 

14.  Suppose  a  man  travel  by  steam  21900  miles  in  a  year ; 
how  far  will  he  travel  in  67  years  ?     Ans.   1467300  miles. 

15.  In  a  volume  of  675  pages,  each  page   containing  156 
lines,  and  each  line    136  letters;  how  many  letters  ?     Ans. 
14320800. 

16.  How  many  hills  are  there  in  a  field  of  corn,  containing 
149  rows,  with  96  hills  in  a  row  ?     Ans.  14304. 

17.  On  the  preceding  supposition,  how  many  ears  of  corn  are 
there  in  the  field,  allowing  the  average  to  be  9  to  a  hill  ?  and 
how  many  kernels  of  corn,  allowing  300  to  an  ear  ?     Ans. 
128736  ears  of  corn,  and  38620800  kernels. 

CASE  4th. — WHEN  THERE  ARE  CYPHERS  ON  THE  RIGHT  HAND 

OF  THE  MULTIPLIER,  OR  MULTIPLICAND,  OR  BOTH. 

RULE. — Omit  the  cyphers  and  multiply  by  the  significant 
figures  only,  and  annex  to  the  right  hand  of  the  product  as 
many  cyphers  as  were  omitted. 

Ex.  1.  Multiply  2100  by  70. 

PERFORMED. 
2100 

70         I  multiply  the  21  by  the  7  only, 
and  then  annex  3  cyphers  to  147t^r 


147000     the  product. 

2.  Multiply  47600  by  150.     Prod.  7140000. 

3.  Multiply  9560  by  1200.     Prod.  11472000, 


SIMPLE  MULTIPLICATION.  41 

4.  Multiply  462000  by  190.     Prod.  87780000. 

5.  Multiply  760  by  1000.     Prod.  760000. 

CASE  5th. — WHEN  THE   MULTIPLIER   is  ANY  NUMBER  BE- 
TWEEN 11  AND  19  INCLUSIVE. 

RULE. — Multiply  by  the  right  hand  figure  only,  and  place 
the  product  under  the  multiplicand  one  place  to  the  right ;  then 
add  it  to  the  multiplicand.  The  sum  will  be  the  true  product. 

Ex.  1.  Multiply  468  by  15. 

PERFORMED. 

46  8x5 

234  0  =  Product  of  468  multiplied  by  5. 

702  0= Product  of  468  multiplied  by  15. 

The  reason  of  this  rule  is  plain.  Were  we  to  multiply  in 
the  ordinary  mode,  the  only  difference  would  be,  that  the  num- 
ber 2340  would  stand  above  468,  instead  of  below  it. 

Ex.  2.  Multiply  37464  by  17. 

PERFORMED. 

3746  4x7 
262248 


63688  8 = Product. 

3.  Multiply  65328  by  13.     Prod.  849264. 

4.  Multiply  23456789  by  14.     Prod.  328395046. 

5.  Multiply  65432  by  15,     Prod.  981480. 

6.  Multiply  123456  by  16.     Prod.  1975296. 

7.  Multiply  437426  by  17.     Prod.  7436242. 

8.  Multiply  653842  by  18.     Prod.     11769156, 

9.  Multiply  603040  by  19.     Prod.  11457760. 
10.  Multiply  999999  by  11.     Prod.  10999989. 

CASE  6th. — WHEN  THE  MULTIPLIER  is  EITHER  21,  31,  41,  51, 
61,  71,  81,  OR  91. 

RULE. — Multiply  by  the  LEFT   hand  figure  only,  and  place 
the  product  under  the  multiplicand  one  place  to  the  left. 

Ex.  1.  Multiply  634982  by  21. 


42  SIMPLE  MULTIPLICATION. 

PERFORMED. 

634  982x2 
1269964=  Product  of  2  tens  one  place  to  the  left. 


13334622  =  Product  of  21. 

2.  Multiply  9382716  by  31.     Prod.  290864196. 

3.  Multiply  1234567  by  41.     Prod.  50617247. 

4.  Multiply  4364369  by  51.     Prod.  222582819. 

5.  Multiply  6937845  by  61.     Prod.  423208545. 

6.  Multiply  364812  by  71.     Prod.  25901652. 

7.  Multiply  482436  by  81.     Prod.  39077316. 

8.  Multiply  2468  by  91.     Prod.     224588. 

Note. — When  in  either  of  the  two  preceding  cases,  cyphers 
intervene  between  the  figures,  the  same  mode  of  operation  may 
be  adopted,  if  care  be  taken  to  give  each  figure  its  true  place. 

Ex.  1.  Multiply  6456  by  105. 

PERFORMED. 
645  6X5 

32280=  Product  of  the  5  placed  two  de- 
grees to  the  right. 
677880=  Product  of  105. 

2.  Multiply  37562  by  601. 

PERFORMED. 

3756  2x6 
225372=  Product  of  5  hundred  placed  two  de- 

grees  to  the  left. 

22574762  =  Product  of  601. 

3.  Multiply  695378  by  5001.     Prod.  3477585378. 

4.  Multiply  2579678  by  1007.     Prod.  2597735746. 

APPLICATION. 

1.  Bought  52  horses   at  $75  each;  what  did  they  cost? 
Ans.  $3900. 

2.  What  cost  84  tons  of  hay  at  $15  per  ton  ?     Ans.  $1260. 

3.  If  a  man  can  travel  43  miles  in  one  day,  how  far  can  he 
travel  in  60  days  ?     Ans.  2580  miles. 

4.  There  are  144  square  inches  in  one  square  foot.     How 
many  square  inches  are  there  in  67  square  feet?     Ans.  9648. 

5.  If  there  be  18  panes  of  glass  in  one  window,  how  many 
are  there  in  a  house  which  has  56  such  windows  ?     Ans.  1008. 

6.  Bought  342  bales  of  linen,  each   containing  56  pieces 
of  25  yards  each.    How  many  yards  did  I  buy  ?  Ans.  478800. 


SIMPLE  DIVISION.  43 

7.  There  is  an  orchard  consisting  of  126  rows  of  trees,  and 
in  each  row  there  are  109  trees.     How  many  apples  are  there 
in  the  orchard,  allowing  an  average  of  1007  on  a  tree  ?     Ans. 
13830138. 

8.  A  certain  State  contains  50  Counties ;  each  County,  35 
towns ;  each  town,  300  houses,  and  each  house,  8  persons. 
What  is  the  population  of  the  State  ?     Ans.  4200000. 

QUESTIONS  —  What  is  the  nature  of  Multiplication  1  How  many 
numbers  are  employed  in  the  operation  1  What  are  they  called,  and 
•what  is  peculiar  to  each  1  What  is  the  number  obtained  called  1  Of 
what  rule  is  multiplication  an  abbreviation  7  Illustrate.  What  are 
the  multiplicand  and  multiplier  called  when  spoken  of  together'? 
What  is  the  value  of  each  figure  in  the  product  when  you  multiply  by 
a  unit  figure  only  1  Units  multiplied  by  units  give  what  7  Units  by- 
tens  7  Units  by  hundreds  1  When  the  multiplying  figure  is  tens, 
what  effect  will  it  have  on  the  value  of  the  product  of  each  figure  in 
the  multiplicand'?  and  what  will  be  the  effect  if  the  multiplying  fig- 
ure be  hundreds  ?  Give  farther  illustration  of  the  value  of  the  pro- 
duct figures.  What  is  case  1st!  What  isihe  rule  7  Case  2d7  The 
rule  7  Case  3d  1  The  rule  1  What  is  a  composite  number  ?  What 
are  the  component  parts  of  a  number  ?  Case  4th  1  The  rule  "?  Case 
5th  1  The  rule  1  Case  Gth  1  The  rule  7 


SIMPLE    DIVISION. 

We  now  come  to  the  reverse  of  the  preceding  rule.  There 
we  had  two  factors  given  to  find  their  product.  Here  we  have 
given  the  product,  or  what  corresponds  to  it,  and  one  of  the 
factors,  and  are  required  to  obtain  the  other  factor.  Multipli- 
cation, as  was  shown,  could  be  performed  by  repeated  addi- 
tions ;  Division  may  be  performed  by  repeated  subtractions. 
Suppose  it  be  required  to  ascertain  how  many  times  4  is  con- 
tained in  12.  It  may  be  done  by  taking  4  from  12,  till  nothing 
remains,  or  till  a  number  less  than  4  remains.  Thus, 
1  2 
4 

8  —  12  —  4. 
4 

4=rl2  —  4  +  4. 
4 

0-12  —  4+4+4. 


44 


SIMPLE  DIVISION. 


The  operation  shows  three  4's  may  be  taken  from  12.  4  is 
therefore  contained  in  12  three  times.  This  is,  however,  a 
slow  mode  of  operation.  A  more  expeditious  one  must  be 
sought ;  and,  preparatory  for  it,  the  scholar  is  required  to  learn 
the  following  table  : 


DIVISION  TABLE. 


2  -4-  2  =  1 


4  -h  2  = 


8  -4-2  = 
10  -4-  2  = 
12  —  2  = 
14 


2  i=  7 
16-2  =  8 
18-2  =  9 
20  -  2  =  10 
22  —  2  =  1 1 
24  -  2  =  12 


3  -f-  3  =  1 
6  -•-  3  =  2 
9  -r-  3  =  3 

12  -  3  =  4 
15  -4-  3  =  5 
18  ~  3  =  6 
21  -4-  3  =  7 
24  -4-  3  =  8 
27  -f-  3  =  9 
30-4-  3  =  10 
33  -T-  3  =  11 
36  -4-  3  =  12 


4  -f-  4  =  1 
8  -4-  4  =  2 
12  -r  4  -  3 
16  -^  4  =  4 
20  -4-  4  =  5 
24  -4-  4  =  6 
28  H-  4  =  7 
32  -r-  4  =  8 
36  -4-  4  —  9 
40  -4-  4  =  10 
44  -f-  4  =  11 
48  -4-  4  =  12 


5  -r-  5  =  1 
10-  5  =  2 
15  -  5  =  3 
20  -  5  =  4 
25  —  5  =  5 
30  -  5  =  6 
35  -  5  =  7 
40  -  5  =  8 
45  -  5  =  9 
50-5  —  10 
55  -5  =  11 
60  -4-  5  =  12 


18  -  6  = 
24-6  = 


6  =  1 
12  —  6  =  2 
3 

4 
30  -  6  =  5 

36  .4-  6  =  6 
42  -4-  6  =  7 
48  -4-  6  =  8 
54  4-  6  =  9 
60  -r-  6  =  10 
66  -4-  6  =  11 
72  -f-  6  =  12 


7-4-7  =  1 
14  _i_  7  —  2 
21  -4-  7  =  3 

28  -f-  7  = 
35  -f-  7  = 
42  -f-  7  = 
49  -f-  7  = 


56  +.  7  =  8 
63  -4-  7  =  9 
70  —  7  =  10 
77  _^  7  =  11 
84  -  7  =  12 


8-4-8=1 
16  -r-  8  =  2 
24  -4-  8  =  3 
32  .-=-  8  —  4 
40  -r-  8  —  5 
48  ~  8  =  6 
56  ~  8  =  7 
64  -f-  8  =  8 
72  —  8  =  9 
80  —  8  =  10 
88  -4-  8  =  11 
96  .i-  8  =  12 


9-^9=  1 
18  -  9  =  2 
27  -f-  9  =  3 

36  -:-  9  =  4 
45  -f-  9  =  5 
54  -f-  9  =  6 
63  ~  9  =  7 
72  -:-  9  =  8 
81  -f-  9  =  9 
90  -4-  9  =  10 
99  -r-  9  =  11 
108  -^  9  =  12 


10 

20 

30 

40 

50 

60 

70 

80 

90 

100 

110 

120 


10  =  1 
10  =  2 
10  =  3 
10=  4 
10=  5 
10=  6 
10=  7 
10=  8 
10  =  9 
10  =  10 
10=  11 
10=  12 


11  -f-  11  =  1 

22  -4-  11  =  2 

•33  -4-  11  =  3 

44  -4-  11  =  4 

55  -4-  11  =  5 

66  -4-  1 1  =  6 

77  -4-  1 1  =  7 

88  -  11  =  8 

99  -  11  =  9 

110-  11  =  10 

121  -  11  =  11 

132  -  11  =  12 


12  -  12  =  1 

24  —  12  =  2 

36  -  12  =  3 

48  ~  12  =  4 

60  ~  ]2  =  5 

72  -4-  12  =  6 

84  ~  12  =  7 

96  -4-  12  =  8 

108  -4-  12  =  9 

120  -4-  12  =  10 

132  -4-  12  =  11 

144  -^  12  =  12 


SIMPLE   DIVISION.  45 

MENTAL  EXERCISES  IN  DIVISION. 

Bought  4  apples  for  8  cents  ;  what  was  the  price  of  one  ? 
8^-4=how  many?  Bought  3  oranges  for  12  cents  ;  what  was 
the  price  of  one  ?  12-^-3= how  many?  At  another  time 
bought  5  for  15  cents ;  what  was  the  price  ?  15-^-5z=how  ma- 
ny? Paid  24  cents  for  6  peaches  ;  what  did  one  cost  ?  24-^-6 
=how  many  ?  What  is  the  cost  of  one  orange  when  8  cost 
24  cents  ?  when  4  cost  24  cents  ?  when  3  cost  24  cents  ? 
24-h8=rhow  many  ?  24-4-4=how  many?  24n-3r=how  ma- 
ny ?  Paid  35  cents  for  7  melons  ;  what  was  the  value  of  one? 
Paid  the  same  money  for  5  ;  what  was  the  value  of  one  ?  35  H- 
7=how  many?  35-^5z=how  many?  Bought  8  silks  for  16 
cents  ;  what  was  one  worth  ?  What  is  the  value  of  one  when 
I  pay  24  cents  for  8  ?  when  I  pay  32  cents  ?  40  cents  ?  48 
cents?  56  cents  ?  16-^-8— how  many  ?  2  4  -=-  8= how  many  ? 
32-?-8=:how  many?  40-h8=how  many?  4 8 H- 8= how  ma- 
ny ?  56-^8= how  many?  64  — 8— how  many?  72-^8  = 
how  many?  Paid  18  shillings  for  9  yards  of  calico;  what 
was  the  price  per  yard  ?  What  is  the  price  per  yard  when 
the  same  quantity  costs  27  shillings  ?  45  shillings  ?  36  shil- 
lings ?  72  shillings?  63  shillings  ?  1 8 ^-9  =  how  many  ?  27 -f- 
9=how  many?  45-f-9=:how  many?  36-^9z=how  many? 
72  +  9= how  many  ?  63  -4-  9 = how  many  ?  What  is  the  cost 
of  one  lemon,  if  10  cost  30  cents?  70  cents  ?  60  cents  ?  50 
cents  ?  80  cents  ?  40  cents  ?  90  cents  ?  30^-  I0=how  many  ? 
&c.  What  is  the  cost  of  one  sheep,  when  8  cost  24  dollars  ? 
What,  if  8  cost  64  dollars?  88  dollars?  72  dollars?  48 
dollars?  80  dollars?  40  dollars  ?  32  dollars?  24-^-8=how 
many?  6  4  H-  8=  how  many  ?  &c. 

What  is  the  value  of  one  bushel  of  wheat,  when  12  bushels 
are  worth  24  dollars  ?  What,  when  the  12  are  worth  36  dol- 
lars ?  72  dollars?  96  dollars  ?  48  dollars  ?  108  dollars? 
60  dollars?  132  dollars?  84  dollars?  120  dollars?  144 
dollars?  24^-12=how  many?  36-^  12:=  how  many?  &c. 
Bought  8  pounds  of  raisins  for  72  cents ;  what  was  the  price 
per  pound?  If  9  pounds  of  rice  cost  81  cents,  what  will  one 
pound  cost  ?  A  man  traveled  96  miles  in  8  days  ;  how  far 
was  that  per  day  ?  A  horse  ran  54  miles  in  6  hours  ;  how  far 
was  that  per  hour  ?  How  many  pounds  of  sugar  at  9  pence 
per  Ib.  can  be  bought  for  54  pence  ?  for  63  pence  ?  for  81 
pence  ?  for  108  pence  ?  for  99  pence  ?  for  72  pence  ?  for  45 
pence  ?  for  36  pence  ? 


46  SIMPLE  DIVISION. 

The  scholar  must  now  be  prepared  to  apply  what  he  has 
learnt  from  the  preceding  table  and  questions,  to  division  on  a 
more  extensive  scale.  He  will  have  noticed  that  for  each  op- 
eration two  numbers  are  given ;  viz.  a  number  to  be  divided, 
which  is  called  the  dividend ;  and  a  number  by  which  to  di- 
vide, called  the  divisor.  The  number  obtained  is  called  the 
quotient;  a  word  which  signifies,  how  many;  because  this 
number  always  determines  how  many  times  the  divisor  is 
contained  in  the  dividend.  After  the  operation  is  performed 
there  is  frequently  a  number  left.  This  is  called  the  remain- 
der, and  is  always  less  than  the  divisor.  When  the  division 
is  performed,  if  there  be  no  remainder,  the  quotient  multiplied 
by  the  divisor  will  always  produce  the  dividend ;  and  if  there 
be  a  remainder,  the  dividend  will  be  produced  by  multiplying 
as  before,  and  adding  the  remainder  to  the  product.  Hence 
division  is  proved  by  multiplication.  The  scholar  will  readily 
perceive  that  these  two  rules  are  the  reverse  of  each  other. 

The  operations  in  division  will  be  illustrated  under  two  gen- 
eral heads ;  viz.  Short  Division,  and  Long  Division. 

I.  SHORT  DIVISION. 

When  the  divisor  does  not  exceed  12,  the  process  is  abbre- 
viated by  keeping  the  computation  in  the  mind,  and  writing 
down  only  the  quotient  figures. 

RULE. — 1st.  Write  down  the  dividend,  and  place  the  divisor 
on  the  left,  with  a  curve  line  drawn  between  them. 

2d.  Take  as  many  figures  on  the  left  of  the  dividend,  as 
will  contain  the  divisor  once  or  more,  and  write  the  figure  ex- 
pressing the  number  of  times,  directly  under  those  divided. 

3d.  If  in  dividing  there  be  a  remainder,  imagine  the  next 
figure  in  the  dividend  to  be  placed  on  the  right  hand  of  it. 
This  will  form  a  new  number,  which  may  be  divided  as  before. 
Continue  the  same  process  till  all  the  figures  of  the  dividend 
have  been  disposed  of,  and  the  number  obtained  will  be  the  quo- 
tient required. 

4th.  If  in  taking  any  figure  of  the  dividend  the  number 
produced  be  not  sufficient  to  contain  the  divisor  once,  a  cypher 
must  be  placed  in  the  quotient,  and  another  figure  of  the  dividend 
taken. 


SIMPLE  DIVISION.  47 

Ex.  1.  Divide  496  by  4. 

PERFORMED. 
4)496 

1  2  4= quotient. 

I  first  say  4  is  in  4  once,  and  place  the  1  as  directed  by  the 
rule.  I  next  say  4  is  in  9  twice  and  1  remains  ;  I  write  down 
the  2  as  directed,  and  on  the  right  hand  of  the  1  I  place  the  6, 
and  thus  obtain  16.  Lastly,  I  say,  4  is  in  16  four  times.  This 
operation  gives  124  as  the  number  of  times  which  496  con- 
tains 4.  Now  the  scholar  will  readily  perceive,  that  this  is  re- 
versing a  process  of  multiplication.  If  he  multiply  124  by  4, 
he  will  obtain  16  units,  8  tens,  and  4  hundreds  ;  and  these  are 
precisely  the  numbers  divided.  But  the  16  units  are  equal  to 
1  ten  and  6  units  ;  the  whole  is,  therefore,  equal  to  4  hundreds, 
9  tens,  and  6  units  ;  or  to  496. 

This  multiplication  is  the  proof  of  the  work. 

Ex.  2.  Divide  1512  by  7. 

PERFORMED. 
7)1512 

2  1  6= quotient. 

The  numbers  here,  as  they  are  severally  divided,  are  15, 11, 
and  42.  If  216  be  multiplied  by  7,  it  will  produce  the  same 
numbers. 

3.  Divide  5463  by  3. 

PERFORMED. 
3)5463 

1821 

4.  Divide  1256  by  2. 

PERFORMED. 
2)1256 

628 

5.  Divide  63548  by  4.   Quo.  15887. 

6.  Divide  256788  by  8.   Quo.  32098  and  4  remains. 

7.  Divide  65342167  by  4  and  by  5,  and  add  the  quotients. 
Ans.  29403974  and  5  remains. 


48  SIMPLE  DIVISION. 

'8.  Divide  735649  by  5  and  by  7,  and  add  the  quotients. 
Ans.  252221  and  9  remains. 

9.  Divide  456789  by   6  and  by  8,  and  add  the  quotients. 
Ans.  133229  and  8  remains. 

10.  Divide  68890  by  7  and  by  9,  and  add  the  quotients. 
Ans.  17495  and  7  remains. 

11.  Divide  78901  by   8  and  by  10,  and  add  the  quotients. 
Ans.  17752  and  6  remains. 

12.  Divide  89012   by  9  and  by  11,  and  add  the  quotients. 
Ans.  17982  and  2  remains. 

13.  Divide   90123456  by  10  and  12,  and  add  the  quotients. 
Ans.  16522633  and  6  remains. 

14.  Nine  persons   drew  a  prize   of  $198;  what  was  each 
one's  share?     Ans.  $22. 

15.  Paid  $750  for  30  cows ;  what  was  the  average  price  ? 
Ans.  $25. 

16.  A  person  dying  leaves  an  estate  of  $4500   to  9  chil- 
dren ;  what  will  be  each  one's  share  ?     Ans.  $500. 

17.  A  man  traveled  224  miles  in  7  days  ;  what  was  his  daily 
progress  ?     Ans.  32  miles. 

18.  If  12  ounces  make  a  pound  of  silver  ;  how  many  pounds 
ajre  there  in  2040  ounces  ?     Ans.  170. 

19.  How  many  times  may  12  be  subtracted  from  1416.     Ans. 
118. 

20.  Four  persons  boarded  at  a  public  house  till  the  bill  of 
their  board  was  $184  ;  what  was  the  average  bill  ?     Ans.  46. 

When  the  divisor  is  more  than  12  and  is  a  composite  num- 
ber, the  same  mode  of  operation  can  be  adopted. 

RULE  2d.— Divide  first  by  one  of  the  component  parts,  and 
the  quotient  arising  from  this  division,  by  the  other. 

The  only  difficulty  which  will  here  present  itself,  will  be  to 
ascertain  the  true  remainder.  The  scholar  needs  only  to  re- 
member, that  if  a  remainder  occur  after  the  first  division  only, 
that  is  the  true  remainder.  If  after  the  second  division  only, 
the  true  remainder  is  obtained  by  multiplying  this  remainder 
by  the  first  divisor.  If  there  be  a  remainder  after  each  divi- 
sion, the  true  remainder  is  found  by  multiplying  the  last  re- 
mainder by  the  first  divisor,  and  adding  the  first  remainder. 

Ex.  1.  Divide  864  by  18.  The  component  parts  of  18  are 
6  and  3,  therefore, 


SIMPLE  DIVISION.  49 

« 

6)864 

3)14  4  =  Quotient  of  864  divided  by  6. 

4  8  =  Quotient  of  144  divided  by  3.     This  is  the 
true  quotient  of  864  —  18. 

Ex.  2.  Divide  162641  by  72.     The  component  parts  are  9 
and  8,  therefore, 

9)162641 
8  )  1  8  0  7  1  and  2  remainder. 

2258  and  7  remainder.    Therefore,  7  X 
9-f-2=the  true  remainder,  viz.  65. 

Ex.  3.  Divide  793  by  18.      The  component  parts  are  6 
and  3. 

6)793 


3)1   32  and  1  remains. 


4  4  and  nothing  remains.     Therefore,  by  rule  2d,  1 
is  the  whole  remainder. 

Ex.  4.  Divide   792  by  16.     Component  parts  of  16  are  4 
and  4. 

4)792 

4)198  and  nothing  remains. 

4  9  and  2  remains  ;  therefore,  by  rule  2d,  4x2 
=  8,  the  true  remainder. 

5.  Divide  2592  by  63.     Quo.   41,  and  9  remains. 

6.  Divide  7776  by  108.     Quo.  72. 

7.  Divide  6750  by  15.     Quo.  450. 

8.  .Divide  437639 by  42.     Quo.  10419,  and  41  remains. 

9.  Divide  738246  by  27.      Quo.  27342,  and  12  remains. 

10.  Divide  60400  by  25.     Quo.  2416. 

11.  Divide  45678  by  16.     Quo.  2854,  and  14  remains. 

12.  Divide  4688  by  48.     Quo.  97,  and  32  remains. 

13.  Divide  347628  by  84.     Quo.  4138,  and  36  remains. 

5 


50  SIMPLE    DIVISION. 


II.  LONG  DIVISION. 

WHEN    THE    DIVISOR    EXCEEDS    12,    AND    IS    NOT    A    COMPOSITE 
NUMBER. 

RULE. — 1st.  Write  down  the  dividend,  and,  drawing  a  curve 
line  both  on  the  right  and  left  of  it,  place  the  divisor  on  the 
left. 

2d.  Find  how  many  times  the  divisor  is  contained  in  the 
fewest  figures  that  will  contain  it,  taken  from  the  left  of  the  div- 
idend ;  and  place  the  figure  expressing  the  number  of  times  on 
the  right  of  the  dividend,  as  the  first  quotient  figure. 

3d.  Multiply  the  divisor  by  this  quotient  figure,  and  place  the 
product  under  the  figures  divided. 

4th.  Subtract,  and  to  the  remainder  bring  down  the  next  figure 
of  the  dividend. 

5th.  Divide  again  and  place  the  result  as  the  second  figure  in 
the  quotient. 

6th.  Continue  the  process  of  multiplying,  subtracting,  bring- 
ing down,  <$fc.  till  all  the  figures  have  been  divided. 

7th.  If  after  all  the  figures  have  been  divided,  there  be  still  a 
remainder,  place  it  as  the  numerator,  and  the  divisor  as  the  de- 
nominator of  a  fraction,  on  the  right  hand  of  the  quotient. 

Note  1st. — Whenever  a  figure  has  been  placed  on  the  right 
of  the  remainder,  and  the  number  produced  will  not  contain  the 
divisor,  a  cypher  must  be  placed  in  the  quotient. 

Note  3d. — If  the  remainder,  after  subtracting,  be  greater  than 
the  divisor,  the  quotient  figure  is  too  small.  If  the  number  ob- 
tained by  multiplying  the  divisor  by  the  quotient  figure,  be 
greater  than  the  number  divided,  that  quotient  figure  is  too 
large. 

Example  1.  Divide  15359  by  29. 

PERFORMED. 

29)15359(529  if. 
1   4  5 


8  Rem. 


SIMPLE    DIVISION.  51 

Explanation. — I  in  the  first  place  notice  that  at  least  three 
figures  are  required  to  contain  the  divisor,  and  that  in  tl^is 
number,  153,  the  divisor  is  contained  5  times.  5  is  then  my 
first  quotient  figure.  I  then  proceed  to  multiply  and  subtract 
as  the  rule  directs,  and  obtain  a  remainder  of  8.  To  this,  I 
bring  down  the  next  figure  of  the  dividend,  and  obtain  85.  I 
proceed  tojiivide,  multiply,  and  subtract,  as  before,  and  obtain  a 
remainder  of  27,  arid  a  quotient  figure  2.  Again,  I  bring  down, 
divide,  multiply,  &c.  and  obtain  the  quotient  figure,  9,  and  a  re- 
mainder of  18,  which  being  placed  according  to  the  7th  article 
of  the  rule,  gives  the  fraction  Jf . 

Note  3d. — It  will  be  observed  that  each  figure  in  the  quo- 
tient is  obtained  by  four  successive  operations,  and  that  these 
operations  uniformly  succeed  each  other  in  the  same  order. 
In  the  first  place,  we  are  required  to  divide ;  in  the  second,  to 
multiply  the  divisor  by  the  quotient  figure  ;  in  the  third,  to  sub- 
tract the  product  of  this  multiplication  from  the  figures  divided ; 
and,  lastly,  to  the  remainder  thus  obtained,  to  bring  down  another 
figure  of  the  dividend. 

Ex.  2.  Divide  6283459  by  29. 

PERFORMED. 

29)6283459(21667  I  =  Quotient. 
5  8 


0  0 


52  SIMPLE    DIVISION. 

3.  Divide  74^1389  by  95. 

PERFORMED. 

95)7461389(78540  |f. 
665 


8  9 

4.  Divide  1893312  by  2076.     Quo.  912. 

5.  Divide  455678  by  78.     Quo.  5842,  and  2  rem. 

6.  Divide  6495685  by  85.     Quo.  76419,  and  70  rem. 

7.  Divide  9424789962  by  978.     Quo.  9636799,  rem.  540. 

8.  Divide  2686211248  by  296.     Quo.  9075038. 

9.  Divide  84764367  by  431.     Quo.  196669,  rem.  28. 

10.  Divide  4683579  by  234.     Quo.  20015,  rem.  69. 

11.  Divide  3579864  by   135.     Quo.  26517,  rem.  69. 

12.  Divide  1748  by  18.     Quo.  97,  and  2  rem. 

Note  4th. — When  the  divisor  is  10,  100,  1000,  10000,  <fec. 
point  off  as  many  figures  from  the  right  of  the  dividend,  as 
there  are  cyphers  in  the  divisor ;  the  figures  on  the  left  of  the 
point  will  be  the  quotient,  and  those  on  the  right,  the  remainder. 

Ex.  1.  Divide  19375468  by  10000.     Quo.  1937,  rem.  5468. 

2.  Divide  99885566  by  100000.     Quo.  998,  rem.  85566. 

3.  Divide  47429  by  10.     Quo.  4742,  rem.  9. 

4.  Divide  463581  by  100.     Quo.  4635,  rem.  81. 

5.  Divide  618293  by  1000.     Quo.  618,  rem.  293. 

Note  5th. — When  the  divisor  consists  of  a  number  of  figures 
with  cyphers  annexed,  the  cyphers  may  be  cut  off  from  the 
divisor,  and  an  equal  number  of  figures  from  the  right  of  the 
dividend,  and  the  remainder  of  the  dividend  divided  by  the 
significant  figures  of  the  divisor.  After  the  division,  the  figures 
cut  off  from  the  dividend  are  to  be  placed  at  the  right  of  the 
remainder. 


SIMPLE    DIVISION.  53 

Ex.  1.  Divide  36418235700  by  98700.  Quo.  368979,  rem. 
8400. 

2.  Divide  11579112  by  890000.  Quo.  13,  rem.  9112. 

3.  Divide  8317642500  by  814600.  Quo.  10210,  rem.  576500. 

APPLICATION. 

1.  If  246  men  incur  an  expense  of  $175152,  what  is  each 
man's  share  ?     Ans.  $712. 

2.  A  gentleman  left  an  estate  of  $65468  to  6  sons ;  what 
was  each  one's  share?     Ans.  $10911f. 

3.  Twelve  men  own  a  bridge,  for  which  they  annually  re- 
ceive $2352,  in  toll ;  what  is  each  man's  share  ?     Ans.  $196. 

4.  Suppose  7776  peach  trees  to  be  planted  in  108  rows  ; 
how  many  trees  are  there  in  a  row?     Ans.  72. 

5.  If  light  comes  from  the  sun  to  the  earth  in  8  minutes, 
how  far  does  it  travel  in  one  minute,    the    distance    being 
95000000  miles  ?     Ans.  11875000  miles. 

6.  If  a  man  travel  9125  miles  in  a  year,  what  is  his  ave- 
rage daily  progress  ?     Ans.  25  miles. 

7.  If  a  horse  run  288  miles  in  36  hours,  how  far  does  he 
run  in  one  hour  ?     Ans.  8  miles. 

8.  In  437850  yards  of  cloth,  how  many  rolls  of  75   yards 
each  ?     Ans.  5838. 

APPLICATION    OF    THE    PRECEDING    RULES. 

1.  A  farmer  sold  3  yoke  of  oxen  at  $96  each;  12  cows  at 
$24  each  ;  83  sheep  at  $3   a  head  ;  239  bushels  of  wheat  at 
$2  per  bushel,  and  distributed  the  avails  equally  among  his  7 
sons.     What  was  each  one's  share  ?     Ans.  $186|. 

2.  A  man  to  whom  was    entrusted  the  settlement   of  an 
estate,  found  that  the  whole  value  of  the  estate  was  $95688. 
There    were    five    claims    against   the    estate ;    viz.    one    of 
$8672;   another  of   $3421;  a  third  of  $10637;    a  fourth  of 
$356  ;  and  a  fifth  of  $1673.     After  the  payment  of  these  seve- 
ral claims,  the  balance  was  to  be  divided  equally  among  9 
heirs.     What  was  the  share  of  each  ?     Ans.    $7881. 

3.  A  farmer  had  16   calves  worth  5   dollars  per  head;    45 
sheep  worth  $3  per  head ;  and  75  bushels  of  grain  worth  $2 
per  bushel ;  he   gave   the  whole  for  a  horse  worth  $136;  a 
carriage  worth  $195,  and  a  harness  worth  $63.     Did  he  gain 
or  lose,  and  how  much?     Ans.  Gained  $29. 

4.  A  merchant  received  by  boat,  9696  bushels  of  salt,  and 

5* 


54  FEDERAL    MONEY. 

hired  it  carted  16  miles,  at  a  dollar  a  load  of  24  bushels, 
How  much  did  the  cartage  cost  ?     Ans.  $404. 

5.  When  the  dividend  is  290864196  and  the  quotient  9382- 
716,  what  is  the  divisor?     Ans.  31. 

6.  A  man  purchased  a  farm  for  which  he  paid  $18000.     He 
sold  60  acres  for  50  dollars  an  acre,  and  then  the  remainder 
stood  him  at  $75  per  acre.     How  much  land  did  he  purchase  T 
Ans.  260  acres. 

GLuESTioNs. — How  does  division  differ  from  multiplication  1  How 
may  multiplication  be  performed  1  How  may  division  be  performed  ? 
How  many  numbers  are  given  in  division  1  What  are  they,  and  what 
are  they  called  1  What  is  the  number  obtained  by  the  operation  called'? 
What  does  it  signify  1  What  is  the  remainder  1  How  does  it  always 
compare  with  the  divisor  1  How  is  division  proved  1  What  is  short 
division  1  What  is  the  rule  for  it  ?  What  is  the  rule  when  the  divi- 
sor is  more  than  12  and  a  composite  number  1  How  is  the  true 
remainder  obtained,  when  we  divide  by  the  component  parts  of  the  divi- 
sor 7  What  is  long  division  1  What  is  the  rule  1  How  do  you  di- 
vide by  10,  100,  1000.  &c.7  When  the  divisor  consists  of  a  figure 
greater  than  1,  with  cyphers  annexed,  how  do  you  divide  1 


FEDERAL  MONEY. 

Federal  Money  is  the  currency  of  the  United  States,  Its- 
denominations  are  Mills,  Cents,  Dimes,  Dollars,  and  Eaglesy 
increasing  like  simple  numbers  in  a  ten-fold  ratio,  as  represent- 
ed in  the  following  table  ; 

TABLE  OF  FEDERAL  MONEY, 

10  Mills  (marked  m.)  make  one  cent,  marked  c.  or  ct, 
10  Cents  "       one  dime,      "        d. 

10  Dimes  "       one  dollar,     "        $  or  doL 

10  Dollars  "      one  eagle,     "       E. 

The  coins  of  the  United  States  are  of  three  kinds  ;  viz. 
gold,  silver,  and  copper  coins. 


The  gold  coins  are, 

The  Eagle       =$10, 
Half-Eagle       =$5. 
Quarter-Eagle = $2  \. 


The  silver  coins  are, 
The  dollar    =100  cts. 
Half-dollar      =50  cts. 
Quarter-dollar =25  cts. 
Dime  =10  cts. 

Half-dime        =   5  cts. 


FEDERAL     MONEY.  55 

The  6$  cent  and  12£  cent  pieces,  &c.  are  not  American 
coinage. 

The  copper  coins  are,  the  cent  and  the  half-cent.  Half- 
cents  are  seldom  used.  The  gold  and  silver  coins  are  not 
composed  of  pure  metal ;  but  are  alloys, — that  is,  compounds  of 
these  metals  with  the  baser  metals.  The  purity  of  a  metal  is 
expressed  by  the  word,  carat;  which  word  is  used  to  ex- 
press a  twenty-fourth  part  of  a  given  quantity.  If,  for  example, 
a  quantity  of  gold  be  said  to  be  18  carats  fine,  the  meaning  is, 
that  18  equal  portions  of  the  whole  are  gold,  and  6  equal  por- 
tions are  of  a  less  valuable  metal. 

By  recurring  to  the  preceding  table,  it  will  be  seen  that  the 
denominations  of  federal  money  may  be  added,  subtracted, 
multiplied,  and  divided  by  the  same  rules  as  the  simple  num- 
bers, as  they  increase  in  the  same  ratio.  The  scholar  should 
however  remember  that  the  dollar  is  the  unit  money  ;  and  that 
dimes,  cents,  and  mills  are  decimals  ;  or  tenths,  hundredths, 
and  thousandths  of  the  dollar,  or  unit  denomination. 

It  is  therefore  always  important  to  know  which  of  a  number 
of  figures  is  the  dollar,  or  unit  figure.  This  is  shown  by  the 
decimal  point  ( . )  or  period,  which  always  stands  between  the 
dollars  and  dimes ;  thus,  63.78,  read,  sixty-three  dollars  and 
seventy-eight  cents. 

The  first  figure  on  the  left  of  the  point  is  dollars,  and  the 
second  eagles.  They  are  however  called  dollars  indiscrimi- 
nately ;  as  in  the  above  number,  the  6  is  eagles,  and  the  3,  dol- 
lars ;  but  usually  read,  63  dollars.  Likewise,  the  78  is  usually 
read,  78  cents,  instead  of  7  dimes  and  8  cents.  Both  methods 
express  the  same  value. 

To  reduce  the  higher  denominations  to  the  lower,  it  is  neces- 
sary to  bear  in  mind,  1st.  That  cents  are  converted  into  mills 
by  annexing  one  cypher;  thus,  8  cents  =  80  mills.  2d.  That 
dollars  may  be  changed  into  cents  by  annexing  two  cyphers  ; 
thus,  3  dollars  =  300  cents  ;  and  into  mills  by  annexing  three 
cyphers;  thus,  $3  =  3000  mills.  3d.  The  reverse  operation 
will  convert  mills  into  cents  and  cents  into  dollars. 

Ex.  1 .  How  many  mills  in  47  cents  ?     Ans.  470. 

2.  How  many  mills  in  69  cents  ?     Ans.  690. 

3.  How  many  mills  in  156  cents  ?     Ans.  1560. 

4.  In  78  dollars,  how  many  cents  1     Ans.  7800.     How  many 
mills  ?     Ans.  78000. 

5.  In  $637,  how  many  cents  1     Ans.  63700.     How  many 
mills  ?     Ans.  637000. 


56  FEDERAL     MONEY. 

6.  In  450  mills,  how  many  cents  ?     Ans.  45. 

7.  In  470  mills,  how  many  cents  ?     Ans.  47. 

8.  In  6700  mills,  how  many  cents  ?     Ans.  670. 

9.  In  6700  mills,  how  many  dollars  ?     Ans.  67. 

10.  Change  $6  into  cents.     11.  Change  42  cents  into  mills. 
12.  Change  $95  dollars  into  mills.     13.  Change  460  mills  into 
cents.     14.  28000  mills  into  dollars.     ]  5.  439  mills  into  cents. 
16.  9876  mills  into  dollars. 

ADDITION  OF  FEDERAL  MONEY. 

RULE. — Set  the  numbers  one  under  another,  so  that  dollars 
shall  stand  under  dollars,  dimes  under  dimes,  cents  under  cents, 
and  mills  under  mills.  Then  add  up  the  several  columns  as  in 
Simple  Addition,  and  place  the  decimal  point,  in  the  amount, 
directly  under  those  in  the  numbers  added. 

If  the  above  rule  be  followed  in  writing  down  the  several 
numbers,  the  separating  points  will  stand  directly  under  each 
other. 

Ex.  1.  What  is  the  sum  of  136  dollars  21  cents;  75  dol- 
lars 13  cents  ;  7  dollars  78  cents  ;  66  dollars  19  cents  ;  and  196 
dollars  72  cents  ?  Ans.  482.03. 

PERFORMED. 
1    3    6.2    1 

75.13 

7.7  8 

66.19 

196.72 


$482.03  Amount. 

2.  Add  together  $432.73  ;  $297.38  ;  $172.66  ;  and  $333.62. 
Amount,  $1236.39. 

3.  What  is  the   sum  of  $1.55;    $0.72;    $340.89;  $0.01; 
$1460.99?     Ans.  $1804.16. 

4.  What  is  the  sum  of  $72.01 ;  $1 ;  $0.01  ;  $0.10;  $40.70; 
$560.88?     Ans.  $674.70. 

5.  What  is  the  sum  of  $101.01;  $20.15;  $42.89;  $79.81; 
$41.41;  $51.51;  $38.41?     Ans.  $375.19. 

6.  What  is  the  sum  of  $16.64;  $20.84;  $462.573;  $29.922; 
$56.32;  $84.48?     Ans.  $670.775. 

7.  A  farmer  bought  a  cow  for  $23.75;  a  yoke  of  oxen  for 


FEDERAL  MONEY.  57 

$96.78;  a  horse  for  $69.83;  and  a  pig  for  $1.625;  what  did 
they  all  cost  him?     Ans.  $191.985. 

8.  A  grocer  paid  for  a  box  of  cheese,  $37.21 ;  for  candles, 
$8.32;  for  a  cask  of  wine,  $7.38;  for  a  box  of  raisins,  $3.625; 
what  was  the  whole  cost?     Ans.  $56.535. 

9.  Bought  5  gallons  of  melasses  for  $1.80;  three  pounds  of 
tea  for  $1.125;  3  yards  of  broadcloth  for  $9.82;   11  yards  of 
cotton  cloth  for  $1.83  ;  6  yards  of  linen  for  $3.82  ;  what  was 
the  amount  of  my  bill  ?     Ans.  $18.395. 

10.  Paid  for  building  my  house,   $2169.72  ;  for  my  barn, 
$972.87;  for  my  out-houses,  $1272.69;  for  digging  my  well, 
$56.38;  what  is  the  amount  of  my  expenses  ?     Ans.  $4471.66. 

11.  My  expenses  for  a  journey  were  as  follows:  for  stage 
fare,  $66.89;  for  meals,  $18.50;  for  lodging,  $6.58;  for  carry- 
ing baggage,  $2.57;  for  cleansing  boots,  $1.36  ;  for  washing, 
$2.66;  what  is  the  amount?     Ans.  $98.56. 

12.  My  hat  cost  me  $4.75 ;  my  coat,  $18.96  ;  my  pantaloons, 
$9.74;  my  vest,  $5.82  ;  and  my  boots,  $6.54;  what  did  the 
whole  cost?     Ans.  $45.81. 

SUBTRACTION  OF  FEDERAL  MONEY. 

RULE. — Place  the  numbers  as  in  Simple  Subtraction,  and 
after  subtracting,  place  the  separating  point  as  directed  in  Ad- 
dition of  Federal  Money. 

Ex.  1.  From  $463.42,  take  $399.99.     Rem.  $63.43. 

2.  From  $179.364,  take  $88.449.     Rem.  $90.915. 

3.  From  $125,  take  $9.09.     Rem.  $115.91. 

4.  From  $642.99,  take  $99.99.     Rem.  $543. 

5.  From  $127.01,  take  $41.10.     Rem.   $85.91. 

6.  From  $200,  take  $0.90.     Rem.  $199.10. 

7.  From  $2,  take  $0.05.     Rem.  $1.95. 

8.  From  $99,  take  $0.99.     Rem.   $98.01. 

9.  I  have  $473,  and  my  brother  twice  as  much,  wanting  90 
cents.     How  much  has  my  brother?     Ans.   $945.10. 

10.  Having  in  my  possession  $1600,  I  paid  $516.95  fora 
span  of  fine  horses;  $156.55  for  a  carriage;  $221.19  fora 
gold  watch  ;  and  spent  $450.71  in  traveling.     How  much  had 
I  left?     Ans.   $254.60. 

11.  If  a  man  has  an  income  of  $4500,  and  spends  $2461.85, 
how  much  does  he  lay  up  ?     Ans.   $2038.15. 

12.  If  one  man  lays  up  $339.86  in  a  year,  and  another 


58  FEDERAL  MONEY. 

$299. 99,  how  much  does  one  lay  up  more  than  the  other  ? 
Ans.   $39.87. 

13.  Bought  a  farm  for  $3946,  and  sold  a  part  for  $1426.82  ; 
what  did  the  remaining  part  cost  me  ?  Ans.  $2519.18. 

MULTIPLICATION  OF  FEDERAL  MONEY. 

RULE. — Multiply  as  in  simple  numbers.  The  product  will 
always  be  in  the  lowest  denomination  given,  which  may  be  re- 
duced to  dollars,  cents,  and  mills,  by  the  preceding  rules. 

Ex.  1.  What  will  36  yards  of  cloth  cost  at  $4.50  per  yard  ? 
Ans.  $162.00. 

PERFORMED. 
4.5    0 

3  6 


2  7.0  0 
1  .3  5.0 

162.00 

The  given  price  being  450  cents,  the  whole  price  is  16200 
cents,  which  equals  $162.00. 

2.  What  will  29  pairs  of  shoes  cost  at  $1.50  per  pair  ?  Ans. 
$43.50. 

3.  What  will  35  pounds  of  beef  cost  at  8  cents  per  pound  ? 
Ans.   $2.80. 

4.  Bought  280  reams  of  paper  at  $2. 35  per  ream;  what  was 
the  whole  cost  ?     Ans.   $658.00. 

5.  What  cost  600  pounds  of  lard,  at  15   cents  per  pound? 
Ans.   $90.00. 

6.  Bought  15  tons  of  hay  at  $16,42  per  ton;  what  was  the 
whole  cost?     Ans.  $246.30. 

7.  What  cost  349  acres  of  land,  at  $15.49  per  acre  ?     Ans. 
$5406.01. 

8.  Bought  18  yoke  of  oxen  for  $72.50  per  yoke  ;  what  was 
the  whole  cost  ?     Ans.  $1305.00. 

9.  Bought  32  pounds  of  butter  at  20  cents  per  pound ;  45 
pounds  of  loaf  sugar  at  18  cents  per  pound  ;  56  pounds  of  cof- 
fee at  15   cents  per  pound;  26  pounds  of  tea  at  $1.75  per 
pound;  21   cwt.  of  raisins  at   $6.75  per  cwt.  ;  42  barrels  of 
flour  at  $7.50  per  barrel ;  and  29  pairs  of  boots  at  $4.50  per 
pair.     What  did  the  whole  cost  me  ?     Ans.  $655.65. 


FEDERAL  MONEY.  59 


DIVISION  OF  FEDERAL  MONEY. 

Division  of  Federal  Money  is  employed  whenever  the  cost 
of  a  number  of  articles,  as  yards,  pounds,  &c.,  is  given,  and 
the  price  of  one  required. 

RULE. — Divide  the  cost  by  the  number  of  articles^  and  point 
off  as  many  figures  from  the  quotient  for  cents  and  mills,  as 
there  are  in  the  given  sum.  If  dollars  only  be  given,  and  cy- 
phers are  added  to  complete  the  division,  these  cyphers  must  be 
regarded  as  cents  and  mills. 

Ex.  1 .  If  9  pounds  of  butter  cost  $225,  what  is  the  value 
of  one  pound  ?  Ans.  $0.25. 

2.  Sold  69  bushels  of  wheat  for  $625  ;  what  was  the  price 
per  bushel  ?     Ans.  $9.057. 

3.  Paid   $75.00  for  500  Ibs.  of  butter ;  what  was  the  price 
per  pound?     Ans.  $0.15. 

4.  Paid  $311.70  for  15  tons  of  hay;  what  was  the  price 
per  ton?     Ans.  $20.78. 

5.  Paid  $658  for  280  reams  of  paper ;  what  did  I  pay  per 
ream?     Ans.  $2.35. 

6.  Paid  $505.44  for  144  Ibs.  of  tea;  what  was  the  price  of 
one  pound?     Ans.  $3.51. 

7.  Paid    $375  for  50  firkins  of  butter ;  what  was  the  price 
per  firkin?     Ans.  $7.50. 

8.  Paid   $43.79  for  29  pairs  of  boots  ;  what  wras  the  price 
per  pair?     Ans.  $1.51. 

9.  Paid  $2.80  for  35  Ibs.  of  beef;  what  was  the  price  per 
pound?     Ans.  .08. 

APPLICATION    OF  THE    PRECEDING    RULES. 

1.  A  man  dying,  left  an  estate  of  $12000,  which  was  divided 
equally  among  7  children  after  his  wife  had  received  her  third. 
What  was  the  portion  of  the   wife,  and  what  did  each  child 
receive  ?     Ans.  $4000,  wife's  portion  ;  $1 142.857,  each  child's 
portion. 

2.  A  man  settling  with  his  grocer,  finds  himself  charged  with 
15  Ibs.  of  tea  at  75  cts.per  pound  ;  42  pounds  of  brown  sugar 
at   11  cts.  a  pound;  3  barrels  of  flour  at   $7.50  per  barrel;*  8 
gallons  of  lamp  oil  at  $1.25  per  gallon  ;  and  45  pounds  of  ham 


60  FEDERAL  MONEY. 

at  15  cts.  per  pound.     He  is  also  credited  $11.62.     How  much 
does  he  owe  his  grocer?     Ans.  $43.50. 

3.  A  man  sells  a  horse  for  $84  ;  5  cows  for  $25  each  ;  and 
agrees  to  take  80   sheep  in  pay.     How  much  do  the  sheep 
cost  him  per  head  ?     Ans.  $2.61^. 

4.  A  person  agrees  to  furnish  a  grocer  with  56  bushels  of 
rye  at  50  cts.  a  bushel,  and  to  take  his  pay  in  coffee  at  15  cts.  a 
pound.     How  many  pounds  of  coffee  will  he  receive  ?     Ans. 
186  J. 

5.  If  I  pay  $21311  for  844  acres  of  land,  what  do  I  pay  per 
acre  1     Ans.   $25.25. 

6.  Bought  350  yards  of  cloth  at  $3  50  per  yard.     Of  this  I 
sold  200  at  $5.00  per  yard.     How  much  money  did  I  pay  out; 
how  much  did  I  receive ;  how  many  yards  had  I   left ;  and 
how  much  did  it  cost  me  per  yard  ?     Ans.  I  paid  out  $1225  ; 
I  received  $1000  ;   150  yards  were  left;   and  what  remained 
cost  me  $1.50  per  yard. 

7.  A  man  sold  his  farm  for  $8456,  and  his  stock  for  $1560  ; 
at  the  same  time  he  had  on  deposite  in  the  bank,  $872  97.     He 
then  purchased  a  house  in  the  city,  for  which  he  paid  $3845  ; 
he  also  purchased  a  horse  and  carriage  for  $392.53,  and  paid 
up  his  old  debts  to  the  amount  of  $1787.     How  much  money 
had  he  left  ?     Ans.  $4864.44. 

8.  Suppose  the  man  in  the  preceding  sum  had  laid  out  the 
balance  of  his  money  in  wagons,  for  which  he  paid  $72  each  ; 
and  that  he  took  these  wagons  to  the  south  and  sold  them  for 
$97   each.     How  many  wagons  did  he  purchase,  and  how 
much. did  he  gain  by  the  transaction,  allowing  his  expenses  to 
have  been  $297.83?     Ans.  He  purchased  67  wagons,  and  had 
$40.44  left;  and  he  gained  $1377.17. 

9.  On  the  first  of  Jan.   a  spendthrift  was  in  possession  of 
$3860.90  :  after  30  days  he  had  only  $1680  remaining ;  how 
much  had  he  spent  during  the  whole  time ;  and  how  much 
daily?     Ans.  Whole  sum,  $2180.90  ;  daily,  $72.696. 

10.  A  man  bought  20  pounds   of  coffee,  for  15   cents  a 
pound  ;  and  18  pounds  of  sugar,  at  12  cts.  a  pound.     He  paid 
96  cents  in  cash,  and  the  balance  in  butter  at  20  cents  a  pound  ; 
how  much  butter  did  it  take  ?     Ans.  21  pounds. 

11.  How  much  tea  worth  56  cents  a  pound  must  be  given 
for  16  sacks  of  salt,  worth  $2.87  per  sack  ?     Ans.  82  pounds. 

QUESTIONS. — What  is  Federal  Money  1  What  are  its  denomina- 
tions 1  How  do  they  increase  in  value?  Repeat  the  table.  What 
are  the  coins  of  the  United  States  1  What  are  the  gold  coins  ?  What 


CANCELING.  61 

are  their  values'?  What  are  the  silver  coins  1  What  are  their  values'? 
Are  the  four-pence  half-penny  and  the  nine-penny  pieces,  American 
coin'?  What  are  the  copper  coins'?  Are  the  gold  and  silver  coins  pure 
metal'?  What  is  their  composition'?  By  what  word  is  the  purity  of 
a  metal  expressed'?  What  does  that  word  express  1  If  a  quantity  of 
gold  be  said  to  be  18  carats  fine,  what  is  meant?  Suppose  it  be 
said  to  be  22  carats  fine,  what  is  meant  1  How  may  the  denomi- 
nations of  Federal  Money  be  added?  Which  is  the  unit  money'? 
What  are  dimes,  cents,  and  mills  ?  How  is  the  dollar  or  unit  figure 
always  shown'?  How  does  the  period  always  stand  1  What  is  the 
first  figure  on  the  left  of  the  point?  The  second1?  How  are  they 
usually  read  1  How  are  the  figures  on  the  right  of  the  point  read? 
How  are  cents  converted  into  mills  1  How  are  dollars  converted  into 
cents  1  What  is  the  rule  for  the  addition  of  Federal  Money  1  How 
is  the  decimal  point  placed?  What  is  the  rule  for  the  subtraction  of 
Federal  Money?  How  is  the  point  placed?  What  is  the  rule  for  the 
multiplication  of  Federal  Money  ?  How  many  figures  are  cut  off 
from  the  right  of  the  product  1  When  is  division  of  Federal  Money 
employed'?  What  is  the  rule?  How  many  figures  do  you  point  off 
in  the  quotient?  If  dollars  only  are  given  and  cyphers  are  added, 
how  are  they  to  be  regarded  ? 


CANCELING. 

In  all  arithmetical  operations,  it  is  important  not  only  to 
know  how  to  solve  a  proposition,  but  also,  how  it  may  be  done 
most  expeditiously .  The  object  of  this  rule  is  to  acquaint  the 
scholar  with  a  principle  by  which  peculiar  expedition  is  attained 
in  the  solution  of  such  sums  as  involve  in  their  operation  both 
multiplication  and  division.  This  principle  is  founded  on  the 
following  facts : 

First.  The  value  of  any  quotient  depends  on  the  ratio,  or 
relative  size  of  the  numbers  divided ;  that  is,  if  the  dividend 
be  three  times  as  large  as  the  divisor,  the  value  of  the  quotient 
is  3  ;  and  if  it  be  four  times  as  large,  the  value  is  4,  &c. 

Second.  If  two  or  more  numbers  are  to  be  multiplied 
together,  and  their  product  divided  by  any  other  number,  the 
true  result  is  obtained  by  first  dividing  one  of  these  numbers 
by  the  dividing  number,  and  then  multiplying  the  quotient  by 
the  remaining  number  or  numbers.  Thus,  if  it  be  required  to 
multiply  8  by  4  and  to  divide  the  product  by  2,  first  divide  8  by 
2  and  multiply  the  quotient  by  4  ;  thus,  8-^-2=4,  and  4x4=16. 
6 


62  CANCELING. 

The  advantage  of  this  process  will  be  more  obvious  if  we 
take  large  numbers.  Suppose  we  wish  to  multiply  288  by  16, 
and  to  divide  the  product  by  144.  The  usual  process  would 
be  thus : 

288  144)4608(32  quotient. 

16  432 


1728  288 

288  288 

460  8 = product.  000 

But  by  first  dividing,  the  operation  is  much  abbreviated  ;  thus  : 

144)28  8(2 
288 


000  and  16x2  =  32. 

By  the  usual  method,  37  figures  are  required ;  by  the  other, 
only  18.  There  is  still  another  advantage.  The  scholar  can 
see  at  a  glance  that  144  is  contained  in  288,  twice  ;  and  that 
twice  16  is  32  ;  so  that  an  operation,  which  is  long  and  pro- 
tracted, is  often  reduced  nearly  or  quite  to  a  mental  operation. 

Third.  When  any  large  number  is  to  be  divided  by  the 
product  of  two  or  more  smaller  numbers,  it  may  be  divided 
by  each  number  separately.  This  needs  no  explanation  ;  it  is 
the  same  as  dividing  by  the  component  parts  of  any  number, 
instead  of  the  number  itself. 

Fourth.  When  the  operation  is  of  such  a  nature  as  to  re- 
quire the  product  of  several  numbers  to  be  divided  by  the  pro- 
duct of  several  other  numbers,  these  numbers  may  be  divided 
before  multiplication,  and  their  quotients  used  instead  of  the 
numbers  themselves.  For  illustration,  suppose  the  product  of 
36  and  42  is  to  be  divided  by  the  product  of  6  and  7.  The 
usual  mode  of  operation  would  be  as  follows,  viz, : 


1512  dividend. 


CANCELING.  63 

7  x  6=42  divisor  ;    therefore,  4  2  )  1  5  1  2  (  3  6,  the  required 

126  quotient. 

252 

252 


But  by  the  preceding  fourth  principle,  36-^-6  =  6,  and  42 -f- 7  = 
6,  and  6  x  6  =  36  Ans.  In  this  example  the  divisors  are,  as  it 
were,  expunged  or  lost,  since  they  divide  without  remainder. 
This  will  not,  however,  always  be  the  case.  It  will  frequently 
be  necessary  to  assume  some  number,  which  will  divide  some 
two  given  numbers,  without  remainder,  agreeably  to  a  rule  soon 
to  be  given. 

But  for  further  illustration,  suppose  it  be  required  to  multi- 
ply the  numbers  36,  12,  27,  and  72  together,  and  to  divide  the 
product  successively  by  24,  18,  and  48.     Now  it  is  evidently 
desirable  to  arrange  these  numbers  so  that  they  may  be  con- 
veniently compared  with  each  other.    We  will  adopt  the  follow- 
ing mode  : — We  will  place  the  numbers  whose  product  is  to 
form  a  dividend,  above  a  horizontal  line  ;  and  those  whose  pro- 
duct is  to  form  a  divisor,  below  the  same  line,  thus  : 
36.  12. 27. 72  =  Dividend. 
24.  18. 48  =  Divisor. 

Now  by  the  fourth  and  last  principle  laid  down,  I  can  divide  36 
in  the  dividend  and  18  in  the  divisor  by  18,  without  a  remain*- 
der,  and  obtain  2  in  the  dividend  and  1  in  the  divisor ;  thus, 
2   12  27    72 
'g,'      -jg.     I  can  also  divide  72  in  the  dividend  and  24  in 

the  divisor  by  24,  and  obtain  3  in  the  dividend  and  1  in  the  di- 
visor— (it  will  be  remembered  that  the  divisors  stand  below  the 

line)— thus,  ^~~^,  Again,  I  can  divide  12  in  the  divi- 
dend and  48  in  the  divisor  by  12,  and  obtain  1  in  the  divi- 
dend and  4  in  the  divisor  ;  thus,  ^  '  l ''  4>  Again,  I  can  divide 
2  in  the  dividend  and  4  in  the  divisor  by  2,  and  obtain  1  in  the 
dividend  and  2  in  the  divisor  ;  thus,  ~ '-  '  It  is  now  evi- 
dent that  the  division  can  be  carried  no  farther  without  remain- 
der. The  next  step  therefore  is  to  divide  the  product  of  the 
numbers  remaining  above  the  line  by  the  product  of  those  be- 
low it.  The  product  of  those  above  the  line  is  27x3  =  81  ; 
and  of  those  below  the  line,  2;  therefore,  81-^2=40$,  the 


64 


CANCELING. 


number  required.  The  same  result  would  have  been  obtained 
by  multiplying  the  numbers  above  the  line  and  dividing  their 
product  by  the  product  of  those  below  it,  previous  to  canceling. 
In  the  above  example,  as  the  numbers  have  been  canceled, 
they  have  been  omitted,  and  a  new  statement  made.  This  is 
by  no  means  necessary.  One  statement  is  sufficient. 

It  will  be  noticed  that  in  every  instance  division  is  effected 
without  a  remainder.  Such  must  always  be  the  case. 

The  following  rule  will  be  found  a  competent  guide  for  the 
scholar  in  all  operations  of  canceling. 

RULE  1st. — Place  all  the  numbers  whose  product  is  to  form 
a  dividend,  above  a  horizontal  line ;  and  all  those  whose  pro- 
duct is  to  form  a  divisor,  below  the  same  line. 

2d.  Notice  whether  there  are  cyphers  both  above  and  below 
the  line  ;  if  so,  erase  an  equal  number  from  each  side. 

3d.  Notice  whether  the  same  number  stands  both  above  and 
below  the  line  ;  if  so,  erase  them  both. 

4th.  Notice  again  if  any  number  on  either  side  of  the  line 
will  divide  any  number  on  the  opposite  side,  without  remainder  ; 
if  so,  divide  and  erase  the  two  numbers,  retaining  the  quotient 
figure  only  on  the  side  of  the  larger  number. 

5th.  See  if  any  two  numbers,  one  on  each  side,  can  be  di- 
vided by  any  assumed  number,  without  a  remainder ;  if  so,  di- 
vide them  by  that  number,  and  retain  only  their  quotients.  Pro- 
ceed in  the  same  manner  as  far  as  practicable ;  then, 

6th.  Multiply  all  the  numbers  remaining  above  the  line  for 
a  dividend,  and  those  remaining  below,  for  a  divisor. 

7th.  Divide,  and  the  quotient  will  be   the  number  required. 

Note  \  st. — If  only  one  number  remain  on  either  side  of  the 
line,  that  number  is  the  dividend  or  divisor,  according  as  it 
stands  above  or  below  the  line. 

Note  2d. — The  figure  1  is  not  to  be  regarded  in  the  operation, 
because  it  avails  nothing  either  to  multiply  or  divide  by  1 . 

In  the  two  following  examples,  a  separate  statement  will  be 
made  for  each  step  of  the  rule,  as  successively  taken,  with  a 
reference  to  the  rule. 

Ex.  2.  Multiply  together  the  numbers  100,  16,  24,  and  36, 
and  divide  their  product  by  the  product  of  60,  10,  and  16. 


CANCELING.  65 

Statement,    — ^ — JQ  jg  .     By  applying  the  second  step  of  the 

1.  16.  24.  36 

rule,  the  statement  is  reduced  to   — ~ — T~r?-     "Y  applym£ 

o.    J.  lt> 

1        24   36 
the  third  step,  we  obtain  -Lg—    —  •      By  application    of  the 

fourth  step,  we  obtain—1^ — ~ — .     The  statement  is  capable  of 

no  farther  reduction,  for  the  divisors  or  lower  numbers  are  all 

canceled.     The  product  of  the  numbers  remaining  above  the 

line  is,  therefore,  the  quotient  required  ;  viz.  4  X  36  =  144  Ans. 

Ex.  3.  Divide  the  product  of  96,  18,  110,  5,  and  42  by  the 

product  of  50,  27, 11,  and  28.  Statement,  50  27  11  28'  ^ 
second  step  of  the  rule,  we  obtain — '- — :  '  '  -^.  By  the 

Qfi     1  ft  JO  QR    O  49 

third,  ^-gi 1~.     By  the  fifth,  ^-| |j.     9  is   assumed 

for  a  divisor.     By  the  fourth,  — ^ —    — .      By   the  fourth, 

96.  13                                       96    I 
again,     — ^ '-. ;  and  by  the  third,  — ' — '- r.     96  is   the 

O*  1  1 

only  number  remaining  above  the  line,  except  the  1 ,  and  noth- 
ing but  1  remains  below  the  line.  96  is  therefore  the  quotient 
sought. 

To  show  more  completely  the  mode  of  solution,  the  canceled 
numbers  will  be  retained  in  the  following  sum  ;  they  will,  how- 
ever, be  crossed,  to  show  that  they  are  canceled. 

Ex.  4.  Divide  the  product  of  16,  9,  4,  by  the  product  of  3 
and  32. 

3 

Statement,  — -^-OQ-      By  fourth  step   of  the  rule,   -— «-oo* 

3  3*6 

Again,  by  the  same,  — '    '     .     Again,  by  the  same,  *  '    '  *» 

2  2 

and  6x3  =  18,  the  required  quotient. 

The  following  statements  are  solved  without  repetition,  that 
the  scholar  may  obtain  accurate  views  relative  to  the  mode  of 
solution  here  presented. 

5.  Divide  the  product  of  21,  12,  42,  100,  by  the  product  of 
7,  35,  and  50. 
6* 


66  CANCELING. 


States,  ?L££J«;.     Canceled,  *^  Then 

5 

3x6x2xl2=r432,  and  432  —  5  =  86f.  Ans. 

6.  Divide  the  product  of  99,  49,  15,  20,  32,  13,  16  by  the 
product  of  77,  10,  16,  49,  39,  and  12. 

,   99.49.  15.20.32.  13.  16 
Statement,  -^^——-——. 

39          528 

~         ,    ,     99,.  49.  15.  Eft.  S2.  13.  IS 
Canceled,  __ 


7  "SB 

then  3x5x2x8—240,  and  240-f-7=34f.  Ans. 

7.  Divide  the  product  of  164,  88,  47  28,  3,  2,  by  the  product 
of  328,  36,21,  12,32. 

11          4 

Q.  t  4    164.  88.  4.  28.  3.  2     ^         .    ,     UR4.  8&.  4,  SB.  $.  2 

Statement,  —-g.^——.    Canceled,  _-&—  ^^  ; 

29344 

therefore,  11  divided  by  the  product  of  9x3x4  =  108,  is  the 
Ans.;  and  may  be  thus  expressed,  Ty¥.  It  will  be  observed, 
that,  in  the  upper  numbers,  the  product  of  3  and  4  cancels  12 
in  the  lower  numbers  ;  and  generally,  when  the  product  of  any 
two  or  more  numbers  on  one  side  equals  any  number  on  the  op- 
posite side,  they  may  all  be  erased  at  the  same  time.  By  a 
careful  examination  of  the  preceding  seven  examples,  the  scholar 
will  be  prepared  to  solve  the  remaining  examples  without  fur- 
ther elucidation.  He  will,  however,  remember  that  the  only 
object  in  view  is  to  prepare  him  to  make  a  successful  application 
of  the  principle  of  canceling,  to  the  solution  of  useful  problems. 

Note  3d.  —  Whenever  the  product  of  the  remaining  numbers 
above  the  line  is  less  than  the  product  of  those  below,  the  an- 
swer will  be  in  the  form  of  a  fraction,  as  in  the  last  example. 
The  scholar,  when  he  has  learned  respecting  the  nature  of  frac- 
tions, will  find  that  a  number  standing  directly  over  another, 
with  a  line  between  them,  always  indicates  division  ;  that  is, 
the  number  above  the  line  is  divided  by  that  below  it. 

8.  Divide  the  product  of  27,  111,  32,  40,  42,  by  the  product 
of  12,  4,  24,  12,  30.      Quo.  388£. 

9.  Divide  the  product  of  132,  8,  49,  4,  84,  81,  42,  by  the 
product  of  16,  28,  7,  5,  6,  45,  and  35.     Quo. 


COMPOUND  NUMBERS.  67 

10.  Divide  the  product  of  32, 18,   72,  81,  7,  56,  63,  by  the 
product  of  24,  36,  162,  63,  2,  32.      Quo.  147. 

11.  Divide  the  product  of  96,  54,  108,  72,  56,  18,  and  21, 
by  the  product  of  27,  81,  324,  28,  72,  3.      Quo.  199£. 

12.  Divide  the  product  of  16,  27, 42,  44,  55,  and  66,  by  the 
product  of  48,  88,  22,  33,  and  49.      Quo.  19f . 

13.  Divide  6,  8,  10,  5,  20,  25,  36,  48,  and  60,  by  the  pro- 
duct of  9,  12,  15,  28,  36,  42,  and  9.      Quo.  201||f  f 

14.  If  12  horses  eat  15  bushels  of  oats,  how  many  would 
36   horses  eat  in  the  same  time  ?     In  solving  this  sum,  if  I 
divide  15  bushels  by  12,  it  will  evidently  give  me  what  one 
horse  will  eat,  viz.  1|  bushels  ;   and  if  I  multiply  this  by  36, 
I  shall  obtain  the  quantity  that  36  horses  will  eat.     This  sum 
may  then  be  easily  solved  by  the  principle  here  introduced. 

The  statement  would  be  thus  :  — '~. 

3  If 

Solved,^—,  and  15x3=45  bushels. 

But  we  will  not  anticipate  :  enough  has  already  been  done  to 
show  that  the  principle  is  a  practical  one.  We  will  leave  its 
application,  for  future  consideration. 

Q.UESTIONS.—  What  is  the  object  of  the  rule  of  canceling!  What 
problems  may  be  solved  on  this  principle!  What  is  the  first  of  the 
facts  on  which  this  principle  is  founded  !  Illustrate.  What  is  the  sec- 
ond !  Illustrate.  What  is  another  advantage  !  What  is  the  third  fact? 
What  is  the  fourth!  Illustrate.  What  is  the  first  step  of  the  rule!  What 
is  the  second  step  !  The  third  ?  The  fourth  !  The  fifth  !  The  sixth  ! 
The  seventh,  and  last!  What  is  note  first!  What  note  second! 
When  the  product  of  the  remaining  numbers  above  the  line  is  less 
than  the  product  of  those  below,  in  what  form  will  the  answer  be! 
What  does  one  number  standing  directly  over  another,  with  a  line 
between  them,  always  indicate  ! 


COMPOUND   NUMBERS. 

We  have  thus  far  been  operating  with  numbers,  of  the  same 
denomination,  and  increasing  in  the  constant  ratio  of  10. 

There  is,  however,  another  class  of  numbers  composed  of 
several  denominations,  increasing  in  no  uniform  ratio,  and 
requiring  to  be  separately  denoted  or  expressed.  These  are 
called  Compound  Numbers.  Under  this  head  are  included  all 
those  denominations  employed  to  express  measures  of  any  defi- 


68  COMPOUND    NUMBERS. 

nite  kind ;  such  as  length,  breadth,  solidity,  weight,  time,  money, 
capacity,  &c. 

The  following  are  the  tables  of  these  denominations.  They 
require  to  be  made  very  familiar,  as  they  show  how  many  units 
of  each  lower  denomination,  are  equal  to  a  unit  of  the  next 
higher  denomination. 

I.  ENGLISH  MONEY. 

The  denominations  of  English  Money  are  pounds,  shillings, 
pence,  and  farthings. 

TABLE. 

d.         qr. 

4  farthings  (marked  qr.)  make  1  penny,  marked  d.  1=    4. 

12  pence  "      1  shilling,      "      s.  *1=  12=  48. 

20  shillings  "     1  pound,       "      £.       1=20=240=960. 

II.  TROY  WEIGHT. 

By  this  weight,  the  precious  metals,  such  as  gold  and  silver, 
also  jewels  and  precious  stones,  are  weighed.  The  following 
are  the  denominations  : 

TABLE. 

pint.         gr. 

24  grains  (gr.)  make  1  pennyweight,  marked  pwt.  1=    24. 

01. 

20  pennyweights  "     1  ounce,  oz.  1=  20=  480. 

12  ounces  "     1  pound,  "        Ib.       1=12=240=5760. 

III.  AVOIRDUPOIS  WEIGHT. 

By  this  weight,  all  coarse  materials,  such  as  hay  and  grain, 
and  also  the  baser  metals,  such  as  copper,  are  weighed. 

TABLE. 

oz.  dr. 

16  drams  (dr.)  make  1  ounce — oz.  1=        16. 

16  ounces  "     1  pound— Ib.  1=      16=      256. 

28  pounds  "     1  quarter— qr.  1=    28=    443=    7168. 

cwt. 
4  quarters  "      1  hun.  wght.— cwt.   1=  4=  112=  1792=  28672. 

20  hundred         "      1  ton— T.  1=20=80=2240=35840=573440- 

IV.  APOTHECARIES'   WEIGHT. 

This  weight  is  used  by  apothecaries  and  physicians  in  mix- 
ing and  preparing  medicines. 


COMPOUND    NUMBERS.  69 

TABLE. 

3-        gr. 

20  grains  (gr.)  make  1  scruple— 3.  1=    20. 

3  scruples  "     1  dram— 3.  1=    3=    60. 

8  drams  "     1  ounce—?.  1=  8=  24=  480. 

ft. 
12  ounces  "     1  pound— ft.  1—12=96=288=5760. 

V.  CLOTH  MEASURE. 
This  is  the  measure  used  for  measuring  all  kinds  of  cloth. 

TABLE. 

na.      in. 

21  inches  (in.)  make  1  nail— na.  1=  2j. 

qr. 

4  nails  "     1  quarter  yard— qr.  1=  4=  9. 

yd. 

4  quarters  "     1  yard— yd.  1=4=16=36. 

E.  FI. 

3  quarters  "     1  Ell  Flemish— E.  PI.  1=3=12=27. 

E.  E. 

5  quarters  "      1  Ell  English— E.  E.  1=5=20=45. 

6  quarters  "      1  Ell  French— E.  Fr.  "    1=6=24=54. 

VI.  WINE  MEASURE. 

This  measure  is  used  for  measuring  all  liquors,  with  the 
exception  of  beer  and  ale.  The  gallon  contains  231  cubic 
inches. 

TABLE. 

4  gills  (gi.)  make  1  pint — pt.  1=      4. 

qt. 

2  pints  "       1  quart-  qt.  1=    2=      8. 

gal. 

4  quarts  1  gallon— gal.  1  =    4=    8=    32. 

3H  gallons    "       1  barrel— bar.  "1=311=126=252=1008. 

hhd. 
63  gallons     "       1  hogshead— hhd.  1=2=63  =252=504=2016. 

2  hogsheads  "       1  pipe— P. 
2  pipes  "       1  tun — T. 

VII.    ALE  OR  BEER  MEASURE. 

Besides  ale  and  beer,  milk  is  measured  by  this  measure. 
The  gallon  contains  282  cubic  inches. 


70  COMPOUND    NUMBERS. 

TABLE. 

2  pints  (pt.)  make  1  quart— qt.  q\=  P%, 

pal 

4  quarts          "       1  gallon— gal.  1=    4=    8. 

bar. 

36  gallons  1  barrel — bar.  1  =36=144=288. 

54  gallons      "       1  hogshead— hhd.  1=1 1=54=216=432. 

VIII.  DRY  MEASURE. 

This  measure  is  used  for  measuring  all  kinds  of  grain,  fruit, 
salt,  coal,  &c. 

TABLE. 

4  gills  (gi.)  make  1  pint— pt.  P  i=     ^4. 

qt. 

2  pints  "    1  quart — qt.  1==      2=      8. 

pk. 

8  quarts  "    1  peck— pk.  1=      8=     16=    64. 

bu. 

4  pecks  "    1  bushel— bu.  1=    4=    32=    64=  256. 

ch, 

36  bushels         "    1  chaldron— ch.          1=36=144=1152=2304=9216. 


IX.  LONG  MEASURE. 

The  following  denominations  and  numbers  are  used  for  mea- 
suring distance. 

TABLE. 

in.  ba. 

3  barley  corns  (b.  c.)  make  1  inch — in.                                 1=  3. 

ft. 

12  inches  make  1  foot— ft.                                      1=          12=  36. 

yd. 

3  feet  make  1  yard— yd.                             1=        3=          36=  108. 

5i  yards  or  16*  feet  1  rod— rd.  ' 1=  54=  16*=  198=  594. 
40  rods  1  furlong— fur.  T=  40=  220=  660=  7920=  23760. 
8  furlongs  1  mile— m.  1=  8=320=1760=  5280=  63360=  190080. 
3  miles  1  league— L.  1=3=24=960=5280=87120=1045440=3136320. 

60  geographic,  or69j  statute  miles,  make  1  degree  on  the  earth's  sur- 
face.   360  degrees  make  the  earth's  circumference. 


COMPOUND    NUMBERS.  71 

X.  LAND  OR  SQUARE  MEASURE. 

TABLE. 

Sq.  ft.  Sq.  in. 

144  square  inches  (Sq.  in.)  make  1  square  foot — Sq.  ft.      1  =        144. 

s.  Yd. 
9  square  feet  make  1  square  yard — Sq.  yd.  1  =        9  =      1296. 

rd. 

30*  square  yards  1  square  rod— Sq.  rd.       1=    30i=    272*=    39204. 

R. 
40  square  rods— 1  square  rood— R.    1=  40=1210  =10890  =1568160. 

4  square*roods— 1  Acre— A.          1=4=160=4840  =43560  =6272640. 
640  square  acres — 1  square  mile. 

Land  is  usually  measured  by  Gunter's  chain,  which  is  4 
rods  or  66  feet  in  length.  The  whole  chain  is  divided  into 
100  equal  parts,  called  links.  The  link  is  therefore  ^  part  of 
the  rod,  and  is  7T9(fff  inches  in  length.  80  chains,  or  320 
rods,  make  1  mile  in  length.  1  square  chain  makes  1 6  square 
rods,  and  10  square  chains  make  1  acre. 

XI.   SOLID  MEASURE. 

This  measure  is  employed  in  measuring  substances  which 
have  three  dimensions ;  viz.  length,  breadth,  and  thickness. 
Timber,  stone,  &c.,  are  among  these  substances. 

TABLE. 

1728  solid  inches  make 1  solid  foot— S.  ft. 

27  solid  feet  make          1  solid  yard — S.  yd. 

40  feet  of  round  or  50  feet  of  hewn  timber  make  1  ton — T. 

128  sol  id  feet  make 1  cord— C. 

A  pile  of  wood  8  feet  long,  4  feet  wide,  and  4  feet  high,  contains  just 

one  cord,  since  8X4X4  =  128. 

XII.  CIRCULAR  MOTION. 

TABLE. 

60  seconds  (")    make  1  minute — '. 

60  minutes  ('  )        "  1  degree—0. 

30  degrees  "  1  Sign— S. 

12  signs  or  360°      "  1  circle— C. 


72 


REDUCTION. 


XIII.  TIME. 

60  seconds  (sec.)  make  1  minute — m. 
60  minutes  "     1  hour — h. 

24  hours  "      1  day— d. 


1=    24= 


m.  tee. 

1=  60. 

60=  3600. 

1440=  86400. 


7  days  make  1  week—  w.  1=      7=  168=  10080=    604800. 

4  weeks    "     1  month—  m.      "l=  4=    28=  672=  40320=  2419200. 


13  months  1  day  and  6  hrs., 
or  365  days  6  hours,  1 
common  year—  yr. 


1=13=52=365J=8760=525960=31557600. 

TABLE  OF  PARTICULARS. 


12  particular  things  make  1  dozen  —  doz. 


12  dozen 
12  gross 
20  things 
24  sheets 
20  quires 
112  pounds 


gross. 

great  gross. 

score. 

quire  of  paper. 

ream. 

quintal  of  fish. 


REDUCTION  OF  COMPOUND  NUMBERS. 

The  scholar  is  now  requested  to  turn  back  to  the  table  of 
English  Money,  and  from  it  answer  the  following  questions  : 
How  many  farthings  make  a  penny  ?  How  many  make  2  pen- 
nies ?  How  many  make  4  ?  How  many  make  6  ?  How  many 
make  7  ?  How  many  make  8  ?  How  many  9  ?  How  many 
10?  How  many  11  ?  How  many  pence  make  1  shilling? 
How  many  make  2  shillings  ?  How  many  make  3  ?  4  ?  5  ? 
6?  7?  8?  9?  10?  11?  12?  How  many  shillings  make 
1  pound  ?  How  many  make  2  ?  3  ?  4  ?  5  ?  &c.  Which 
is  worth  most,  1  penny  or  4  farthings  ?  2  pence  or  8  far- 
things ?  Which  is  worth  most,  1  shilling  or  12  pence  ?  2 
shillings  or  24  pence  ?  1  pound  or  20  shillings  ?  2  pounds  or 
40  shillings  ?  If  these  expressions  are  equal  in  value,  in  what 
do  they  differ  ?  Ans.  They  are  different  expressions  for  the 
same  value  ?  In  what  then  does  Reduction  consist  ?  Ans.  In 
changing  numbers  from  one  denomination  to  another  without 
altering  the  value.  Reduce  6  pence  to  farthings.  By  what  do 


REDUCTION.  73 

you  multiply  the  6  ?  Ans.  By  4.  And  why  ?  Because  4  far- 
things make  one  penny  ;  and  consequently,  there  must  be  four 
times  as  many  farthings  as  pence  ?  Reduce  4  shillings  to 
pence.  How  do  you  reduce  shillings  to  pence  ?  Reduce  3 
pounds  to  shillings.  How  do  you  reduce  pounds  to  shillings  ? 
In  these  last  examples,  were  high  denominations  brought  into 
low,  or  low  into  high  ?  Ans.  High  denominations  were  brought 
into  low.  How  was  it  effected  ?  Ans.  By  Multiplication. 
Reduce  12  farthings  to  pence.  By  what  do  you  divide  ?  Ans. 
By  4.  Why?  Because  4  farthings  make  one  penny.  Re- 
duce 36  pence  to  shillings.  By  what  do  you  divide  ?  Reduce 
40  shillings  to  pounds  ?  By  what  do  you  divide  ?  What 
change  is  here  made  in  the  denomination  ?  Ans.  Low  de- 
nominations are  brought  into  high.  How  then  are  low  de- 
nominations brought  into  high  ?  Ans.  By  Division.  After  a 
careful  examination  of  the  preceding  questions  and  remarks, 
the  scholar  will  readily  perceive  the  appropriateness  of  the  fol- 
lowing definition. 

Reduction  is  an  operation  by  which  a  number  expressing  the 
value  of  a  quantity  in  one  denomination  is  changed  to  another 
number,  expressing  the  same  value  in  a  different  denomination. 

But  it  has  already  been  shown  that  high  denominations  are 
brought  into  low  by  multiplication,  and  that  low  denominations 
are  brought  into  high  by  division.  The  scholar,  therefore, 
needs  only  the  rules  by  which  to  guide  his  operation. 

WHEN  A  HIGHER  DENOMINATION    IS    TO  BE  REDUCED  TO  A 
LOWER. 

RULE  1st. — Multiply  the  higher  denomination  by  that  num- 
ber which  it  takes  of  the  LOWER  denomination  to  make  ONE  of 
the  higher,  remembering  to  add  to  the  product  whatever  may  be 
given  of  this  lower  denomination.  If  it  be  required  to  reduce 
the  quantity  still  lower,  multiply  the  number  already  obtained 
by  the  number  required  to  reduce  it  to  the  next  lower  denomina- 
tion, adding  to  the  product  the  given  number  of  that  denomina- 
tion, if  any.  Continue  the  same  operation  till  you  come  to  the 
required  denomination. 

WHEN  A  LOWER  DENOMINATION  IS  TO  BE  BROUGHT  TO  A 
HIGHER. 

RULE  2d. — Divide  the  lower  denomination  by  that  number 
which  is  required  of  that  denomination  to  make  one  of  the  next 


74  REDUCTION. 

higher.  The  quotient  obtained  will  be  of  the  higher  denomina- 
tion ;  and  if  there  be  any  remainder,  it  will  be  of  the  same'  de- 
nomination as  the  number  divided.  Divide  the  quotient  again, 
(if  it  be  not  already  reduced  to  as  high  a  denomination  as  prac- 
ticable^) by  the  same  general  principle ;  and  continue  so  to  do 
till  you  have  reached  the  highest  denomination,  of  which  the 
given  quantity  is  susceptible. 

Ex.  1.  Reduce  6  £.  17s.  9d.  3qr.  to  farthings. 

PERFORMED. 

6.     1  7.     9.    3. 
Multiply  by       20 

1  2  Or=  shillings  in  6  pounds. 
1  7= given  shillings  added. 

I  3  7 = whole  number  of  shillings. 
Multiply  by        12 

164  4 = pence  in  137  shillings. 
9 1=  given  pence  added. 

165  3=2 whole  number  of  pence, 
Multiply  by  4 

661  2 = farthings  in  1653  pence. 
3  =  given  farthings  added. 

'661  5  = whole  number  of  farthings. 

Each  pound  is  of  the  value  of  20  shillings  ;  therefore,  6  £. 
=  120  s.  and  6  £.  and  17s.=rl37s.  Each  shilling  is  of  the 
same  value  as  12  pence;  therefore,  137s.  =  1644  d.  and  1644 
d.  +  9d.  =  1653d.  Each  penny  =  4  farthings,  therefore,  1653 
d.=6612qr.  and  6612qf.-f  3qr.=6615qr. 

The  scholar  will  notice  that  each  denomination  below  the 
pounds,  has  been  added  by  a  separate  process.  This  is  not 
necessary  ;  it  may  be  added  mentally,  as  in  the  following  ex- 
ample : 

2.  Reduce  18  £.  13s.  lid.  2qr.  to  farthings.  The  thing 
to  be  done  is  the  same  as  before.  The  scholar  will  bear  in 
mind  that  the  object  of  this  example  is  to  show  that  the  lower 
denominations  may  be  added  mentally. 


REDUCTION.  75 


PERFORMED. 

1  8  £.    13s.    lid.    2  qr. 
2  0 


3  7  3= shillings  in  18  £.  13s.;  the  13s.  being 
1  2  added  mentally. 


448  7=pence  in  373  s.   lid.;  the  11  d.  being 
4  also  added  mentally. 


1795  0= farthings  in  4487  d.  2  qr. ;  the  2  qr.  add- 
ed as  before. 

In  the  above  example,  to  the  product  of  18  multiplied  by 
20,  I  add  1 3  ;  that  is,  I  add  3  to  the  units  and  1  to  the 
tens.  And  when  I  multiply  by  12,  I  say,  12  times  3  shillings 
are  36  pence,  and  the  1 1  given  pence  added  make  47  pence.  I 
proceed  in  the  same  manner  in  reducing  pence  to  farthings. 

The  preceding  examples  will  serve  to  illustrate  Rule  1st. 
An  illustration  or  two  will  also  be  given  of  Rule  2d,  that  is,  of 
bringing  low  denominations  into  high. 

3.  Reduce  17950  farthings  to  pounds,  shillings,  pence,  and 
farthings. 

PERFORMED. 

4)17950 


12)448  7 — 2qr.  =  4487d.  2  qr.  obtained  by  first  division. 
20)37  3—11  d.r=373  s.  11  d.  obtained  by  dividing  by  12. 

1  8 — 13  s™18  pounds  obtained  by  dividing  by  20, 
and  13s.  remain. 

Since  4  farthings  make  1  penny,  it  is  evident  that  there  are 
as  many  pence  in  17950  qrs.  as  there  are  4's  contained  in  it. 
The  same  reasoning  may  be  applied  to  the  other  divisors. 

It  will  be  observed,  that  in  this  last  example  we  have  reversed 
what  was  done  in  the  second  example.  We  there  had  the 
same  value  given,  which  we  have  here  ;  but  were  required  to 
change  it  from  a  higher  to  a  lower  denomination,  instead  of 
from  a  lower  to  a  higher,  as  in  the  last  example. 

4.  Reduce  44447  farthings  to  pence,  shillings,  and  pounds. 


76  REDUCTION. 

PERFORMED. 

4)44447 


12)1111  1 — 3  qr.  remain. 

20)92  5 — 1 1  pence  remain. 

4  6 — 5  shillings  remain. 
The  answer,  therefore,  is  46  £.  5  s.  1 1  d.  3  qr. 

5.  Reduce  22685  qr.  to  pounds,  &c. 

PERFORMED. 

4)22  685 

12)567  1 — 1  qr.  remains. 
20)47  2 — 7  pence  remain. 

2  3 — 12  s.  remain. 
The  answer  is,  23  £.  12  s.  7  d.  1  qr. 

6.  Reduce  7195  pence  to  pounds,  &c. 

The  scholar  will  observe  that  the  given  number  is  already 
pence. 

PERFORMED. 

1  2)7  1  9  5 

20)59  9—7  d.  remain. 

2  9 — 19s.  remain. 
The  answer  then,  is,  29  £.  19  s.  7  d. 

The  scholar  must  first  consider  whether  the  quantity  given 
is  to  be  brought  from  a  higher  denomination  to  a  lower,  or 
from  a  lower  denomination  to  a  higher.  When  this  is  deter- 
mined, let  him  apply  the  corresponding  rule. 

TABLE  I. ENGLISH  MONEY. 

6.  Reduce  23  £.  12s,  7d.  1  qr.  to  farthings.     Ans,  22685 
qr. 

N.  B.  23  £.  12s.  7d.  1  qr.=22685qr.  So  in  all  opera- 
tions of  mere  reduction,  the  quantity  given  equals  in  value  the 
quantity  obtained 

7.  Reduce  71  £.  13s,  2  d.  3qr,  to  farthings,      Ans.  68795 
qr. 


REDUCTION.  77 

8.  Reduce  299924  qr.  to  pounds,  shillings,  &c.  Ans.  312  £. 
8  s.  5  d. 

9.  Reduce  68795  qr.  to  pounds,  shillings,  &c.  Ans.  71  £. 
13s.  2d.  3qr. 

10.  Reduce  46  £.  5  s.  11  d.  3  qr.  to  farthings.     Ans.  44447. 

11.  Reduce  29  £.  19s.  7d.  to  pence  and  farthings.     Ans. 
7195d. ;  28780  qr. 

12.  Reduce  5974681369  qr.  to  pounds,  &c.     Ans.  6223626 
£.  8  s.  6  d  .  1  qr. 

13.  Reduce  7195  pence  to  pounds,  &c.  Ans.  29  £.  19s.  7d. 

14.  Reduce  40320  half-pence  to  pounds.     Ans.  84  £. 

15.  Reduce   125  £.  19s.   lid.  3qr.    to    farthings.      Ans. 
120959  qr. 

16.  Reduce  475  dollars,  at  6  shillings  each,  to  pence.    Ans. 
34200  pence. 

17.  Reduce  312  £.  8  s.  5  d.  to  half-pence.     Ans.    149962 
half-pence. 

18.  Reduce  121  pistoles,  at  22  shillings  each,  to  pence  and 
farthings.     Ans.  31944d. ;   1 27776  qr. 

19.  Reduce   34200  pence  to  dollars,  at  6   shillings  each. 
Ans.  475  dollars. 

20.  Reduce  359548  qr.  to  pistoles,  at  22  shillings    each. 
Ans.  340  pistoles,  10s.  7  d. 

21.  Reduce  740  dollars,  at  8  shillings  each,  to  pence.    Ans. 
71040  pence. 

22.  Reduce  79  £.  to  pence  and  half-pence.     Ans.  18960  d. ; 
37920  half-pence. 

APPLICATION  OF  TABLE  II. TROY  WEIGHT. 

Ex.  1.  Reduce  23  Ib.  9oz.  6pwt.  22  gr.  to  grains. 

PERFORMED. 

2  3.     9.     6.     2  2. 
1  2  oz.  =  l  Ib. 

2  8  5— ounces  in  23  Ib.  9  oz. 
2  Opwt.^1  oz. 

570  6= pwt.  in  285  oz.  and  6  pwt. 
2  4  gr.=:l  pwt. 


22846 
11412 

Ans.  13696  6  =  grains  in  5706  pwt.  and  22  gr. 


78  REDUCTION. 

2.  Reduce  8578  grains  to  pounds,  &c, 

PERFORMED. 

24)8578 

20)35  7 — lOgr.  remain. 
12)1  7 — 17pwt.  remain. 

1 — 5  oz.  remain. 
Ans.  I  Ib.  5oz.  17pwt.  lOgr. 

3.  Reduce  62  Ib.  7oz.  14pwt.  18gr.  to  grains.    Ans.  360834 

S1"- 

4.  Reduce  360834  grains  to  pounds.     Ans.  62  Ib.  7oz.  14 

pwt.  18gr. 

5.  Reduce  1  Ib.  11  oz.  19  pwt.  23  gr.  to  grains.  Ans.  11519 

gr- 

6.  Reduce  11519  grains  to  pounds,  <fcc.     Ans.  lib.  11  oz. 

19  pwt.  23  grains. 

APPLICATION  OF  TABLE  III. AVOIRDUPOIS  WEIGHT. 

Ex.  1.  Reduce  4  cwt.  3  qr.  26  Ib.  10  oz.  12  dr.  to  drams. 

PERFORMED. 

4.     3.     26.     10.     1  2. 

4  Multiply  by  4,  because  4  quarters  =  1  cwt. 

1  9 

2  8  Multiply  by  28,  because  28  pounds  =  1  qr. 

1  78 
3  8 

558 
1  6  Multiply  by  16,  because  16  oz.^1  Ib. 


8938 

1  6  Multiply  by  16  again,  because  16  dr.  =  l  oz. 


53640 
8938 

143020 


REDUCTION.  79 

The  same  reversed : 

1  6)1  43020 

1  6)893  8 — 12  dr.  remain. 
28)558— lOoz.       " 
4)19— 26  Ib.        " 

4_3  qr.          « 
Ans.  4  cwt.  3  qr.  26  Ib.  10  oz.  12  dr. 

2.  Reduce  7  cwt.  3  qr.  19  Ib.  to  ounces.  Ans.  14192  ounces. 

3.  Reduce  14192  ounces  to  pounds,  quarters,   &c.     Ans. 
7  cwt.  3qr.  191b. 

4.  Reduce  29548  ounces  to  hundreds,  quarters,  &c.     Ans. 
16  cwt.  1  qr.  26  Ib.  12  oz. 

5.  Reduce  480  drams  to  pounds,  &c.     Ans.  1  Ib.  14oz. 

6.  Reduce  12  tons  to  ounces.     Ans.  430080  oz. 

7.  Reduce  430080  ounces  to  tons.     Ans.  12  tons. 

APPLICATION  OF  TABLE  IV. APOTHECARIES*  WEIGHT. 

Ex.  1.  Reduce  8  ft.  8 1.  5  3.  2  3.  12  gr.  to  grains. 

8.     8.     5.     2.     1  2. 
1  2  oz.=llb. 

1  04 


837 

33,=  13 

2513 

20gr.  = 

50272  Ans. 
The  same  reversed  : 

20)50272 


3)251  3—  12  remain. 

8)83  7  —  2    " 
12)10  4  —  5    " 

8—8 


oU  REDUCTION. 

3.  Reduce  27  ib.  9  §.  6  3.  2  3.  to   scruples.      Ans.  8012 
scruples. 

4.  Reduce  477816  grains  to  pounds,  &c.     Ans.  82  Jb.  11  f. 
33.  13.  16  gr. 

5.  Reduce  6  ft.  7  §.  6  3.  13.  12  gr.  to  grains.    Ans.  38312 
grains. 

6.  Reduce  6348  scruples  to  pounds.     Ans.  22  ib.  4  3. 

7.  Reduce  480  drams  to  pounds.     Ans.  5  ft>. 

8.  Reduce  1  ft>.  1  §.  1  3.  1  3.  1  gr.  to  grains.      Arcs,    6321 
grains. 

APPLICATION  OF  TABLE  V. CLOTH  MEASURE. 

Ex.  1.  Reduce  30  yd.  3  qr.  3na.  to  nails. 

PERFORMED. 

30.     3.     3. 

4  =  lyd. 

1  2  3 

4  =  lqr. 

495 

The  same  reversed :  495  nails =how  many  qr  ? 
4)4  9  5 

4)12  3 — 3  na.  remain. 
3  0  yd.— 3  qr.  " 

2.  Reduce  450  E.  E.  3  qr.  2  na.  to  nails.     Ans.  9014  nails. 

3.  Reduce  678  ells  Flemish  to  nails.     Ans.  8136  nails. 

4.  Reduce  8136  nails  to  ells  Flemish.     Ans.  678  ells. 

5.  Reduce  9014   nails  to  ells   English.     Ans.  450  E.    E. 
3  qr.  2  na. 

6.  Reduce  12  ells  French,  5  qr.  3  na.  to  nails.      Ans.    311 
nails. 

7.  Reduce  622  nails  to  ells  French.     Ans.  25  ells,  5  qr.  2  na. 

APPLICATION    OF    TABLE    VI, WINE    MEASURE. 

Ex.  1.  Reduce  3  hhd.  42  gal.  2  qt.  1  pt.  to  pints. 


REDUCTION.  81 

• 
PERFORMED. 

3.       42.       2.       1. 

6  3 

1  1 

2  2 

2  3  1 
4 

926 
2 

1853  Ans. 
The  same  reversed  :   2)1853 

4)92  6 — 1  remains. 
63)23  1—2 

3 — 42      " 
Ans.  3  hhd.  42  gal.  2  qt.  1  pt. 

2.  Reduce  6  pipes,  1   hogshead,  and  1   gill,  to  gills.     Ans. 
26209  gills. 

3.  Reduce  30000  gills  to  pipes.     Ans.  7  pipes,  55  gal.  2  qt. 

4.  Reduce  20  tuns  to  gills.     Ans.  161280  gills. 

5.  Reduce   5  hhd.   56  gal.  2  qt.  to  pints.     Ans.  2972  pints. 

6.  Reduce  16  barrels,  21  gal.  to  quarts.     Ans.  2100. 

APPLICATION    OF    TABLE    VII. ALE    OR    BEER  MEASURE. 

1.  Reduce  4  hhd.  45  gal.  3  qt.  to  pints. 

4.     45.     3. 
5  4 


2094     Ans, 


82  REDUCTION. 

The  same  reversed :     2)2094 
4)1047 

54)26  1 — 3  qt.  remains. 
4 — 45  gal.    " 

2.  Reduce  47  barrels  of  beer  to  pints.     Ans.  13536  pints. 

3.  Reduce  13672  pt.  to  barrels,  &c.     Ans.  47  bar.  17  gal. 

4.  Reduce  451  bar.  7  gal.  to  quarts.     Ans.  64972  quarts. 

5.  Reduce  21  hhd.  to  quarts.     Ans.  4536  quarts. 

6.  Reduce  6562  pints  to  hogsheads.     Ans.  15  hogsheads, 
10  gal.  Iqt. 

APPLICATION  OF    TABLE    VIII. DRY    MEASURE. 

Ex.  1.  Reduce  6  ch.  9  bu.  3  pk.  to  gills.     Ans.  57792  gills. 

2.  Reduce  87762  gills  to  chaldrons.     Ans.    9ch.   18  bu.  3 
pk.  2  qt.  2  gi. 

3.  In  1 36  bushels,  how  many  pecks,  quarts,  and  pints  ?     Ans. 
544  pk.  4352  qt.  8704  pt. 

4.  Reduce  10640  pints  to  bushels.     Ans.  166  bu.  1  pk. 

5.  Reduce  3  pecks  to  gills.     Ans.  192  gills. 

6.  Reduce  720  quarts  to  bushels.     Ans.  22  bu.  2  pk. 

APPLICATION    OF    TABLE    IX. LONG    MEASURE. 

1.  Reduce  8  leagues,  2  miles,  6  furlongs,  16  rods,  3  yards, 
2  feet,  9  inches,  and  2  barley  corns,  to  barley  corns.     Ans. 
5094569  b.  c. 

2.  How  many  barley  corns  will  reach  round  the  earth,  it 
being  360  degrees  ?     Ans.  4755801600. 

3.  Reduce   48765000  barley  corns  to  miles,   &c.      Ans, 
256  m.  4  fur.  15  rods,  15  ft.  10  inches. 

4.  Reduce  26431  rods  to  miles,  &c.     Ans.  82m.  4  fur.  31 
rods. 

5.  Reduce  1710720  inches  to  miles.     Ans.  27  miles. 

6.  Reduce  7  fur.  36  rods,  and  9  ft.  to  inches.     Ans.  62676. 

APPLICATION    OF    TABLE  X. LAND    OR  SQUARE  MEASURE. 

1.  Reduce  500  acres  to  square  rods.     Ans.  80000  rods. 

2.  Reduce  32000  rods  or  poles  to  acres.     Ans.  200  acres. 

3.  Reduce  3  square  miles  to  square  rods.     Ans.  307200. 

4.  Reduce  458000  square  rods  to  square  miles,  &c.     Ans. 
4  sq.  m.  302  acres,  2  roods. 


REDUCTION.  83 

5.  Reduce  6272640  square  inches  to  acres.     Ans.  1  acre. 

Note. — For  figures  to  illustrate  square   and  solid  measure, 
the  scholar  is  referred  to  square  and  cube  root. 

APPLICATION    OF    TABLE    XIj — SOLID    MEASURE. 

1.  In  a  pile  of  wood  containing  2  cords  and  64  feet,  how 
many  solid  inches  ?     Ans.  552960. 

2.  Reduce  884736  solid  inches  of  wood  to  cords.     Ans.  4 
cords. 

3.  Reduce  6117120  cubic  inches  to  tons  of  round  timber. 
Ans.  88  tons,  20  ft. 

4.  Reduce  72  tons  of  hewn  timber  to  cubic  inches.     Ans. 
6220800. 

APPLICATION    OF    TABLE    XII. CIRCULAR    MOTION. 

1.  Reduce   6   signs,  21    degrees,  40  minutes,  to  seconds. 
Ans.  726000  seconds. 

2.  Reduce  432000  seconds  to  signs.     Ans.  4   signs. 

3.  Reduce  1    circle,  6  signs,  25  degrees,  to  minutes.     Ans. 
33900  m. 

4.  Reduce  45200  minutes  to  circles,  &c.     Ans.  2  cir.  1  S. 
3°.  20'. 

It  is  desirable  that  the  scholar 
should  obtain  correct  views  of 
this  measure.  The  adjoining 
figure  will  illustrate  its  applica- 
tion. The  circle  is  regarded  as 
an  integral  object.  Ks  first  divi- 
sion is  into  12  signs.  These  are 
represented  by  the  figures,  1 ,  2, 
3,  4,  &c.  Each  sign  is  divided 
into  30  equal  parts  :  these  con- 
stitute degrees ;  and  as  the  circle 
contains  12  signs,  it  is  evident  the  whole  circle  must  contain 
360  degrees.  These  again  are  divided  into  minutes,  and  the 
minutes  into  seconds.  The  degrees,  minutes,  and  seconds  are 
not  represented  in  the  figure.  This  measure  is  obviously  ap- 
plied to  bodies  moving  in  a  circle  ;  such  as  all  wheels  in  ma- 
chinery, the  revolutions  of  the  heavenly  bodies,  &c. 


84 


REDUCTION. 


APPLICATION    OF    TABLE    XIII. TIME. 


1.  Reduce  360  years,  300  days,  20  hours,  50  minutes,  and 
37  seconds,  to  seconds.     Ans.  11386731037. 

In  the  above  sum,  365  days  and  6  hours  are  allowed  to  the 
year. 

2.  Reduce  662709600  seconds  to  years,   &c.  allowing  the 
year  to  be  as  above.     Ans.  21  years. 

3.  Reduce  49  weeks  to  seconds.     Ans,  29635200  sec. 

4.  Reduce  59270400  seconds  to  years,   &c.     Ans.  I  year, 
46  weeks. 

5.  Suppose  my  age  to  be  21  years,  how  many  seconds  have 
I  lived?     Ans.  662709600. 

6.  Reduce   1325419200  seconds  to  hours.     Ans.  368172 
hours. 

The  following  particulars  require  to  be  introduced  here. 
The  year,  as  given  above,  contains  365  days  and  6  hours.  In 
four  years,  this  six  hours  amounts  to  24  hours,  or  one  day. 
Hence  every  fourth  year  has  366  days.  This  is  called  Bis- 
sextile or  Leap  year.  As  it  requires  four  years  to  gain  this 
one  day,  it  is  obvious  that  the  Leap  year  may  be  found  by  divi- 
ding the  given  year  by  4.  If  it  be  Leap  year  it  will  divide 
without  remainder.  If  in  dividing  there  be  1  remaining,  the 
year  under  consideration  is  the  first  after  Leap  year.  If 
there  be  2  remaining,  it  is  the  second,  and  if  3  remain,  it  is  the 
third  after  Leap  year.  Thus  1824,  is  Leap  year,  for  1824-^4 
—  456,  and.  no  thing  remains.  The  same  operation  shows  1826 
to  be  the  second  after  Leap  year.  In  the  table  the  year  is  divi- 
ded into  13  months.  These  are  lunar  months.  It  is  much 
more  usually  divided  into  12  calender  months,  containing  each 
the  following  number  of  days,  viz.  April,  June,  September,  and 
November,  30 ;  January,  March,  May,  July,  August,  October, 
December,  31  ;  and  February,  28.  To  this  last  month,  (Febru- 
ary,) the  additional  day  of  the  Leap  year  is  added,  so  that 
every  fourth  year,  this  month  has  29  days. 

PROMISCUOUS    EXAMPLES. 

1.  In  64126  gills,  how  many  bushels  ?     Ans.  250  bu.  1  pk. 
7qt.  1  pt.2gi. 

2.  In  26709912  barley  corns,  how  many  leagues  1     Ans. 
46  1.  2m.  4  fur.  6  rods,  1  yd. 

3.  In  161280  gills,  how  many  tuns  of  wine  ?     Ans.  20. 


REDUCTION,  85 

4.  In  10  cords  of  wood,  how  many  solid  inches?     Ans. 
2211840. 

5.  In  20  hhd,  of  sugar,  each   12  cwt.  how  many  pounds  ? 
Ans.  26880. 

6.  How  many  gills  in  250  bu.  1  pk.  7  qt.  1  pt.  2  gills  ?    Ans. 
64126. 

7.  How  many  pence  are  there  in  16  bags,  containing  each 
24  guineas,  16  shillings,  and  8  pence,  the  guinea  being  28s.? 
Ans.   132224. 

8.  In  15840  yards,  how  many  leagues  ?     Ans.  3. 

9.  In  1876742  solid  inches,  how  many  cords,  &c.  ?     Ans. 
8  cords,  62ft.  134  in. 

Thus  far,  examples  have  been  avoided,  which  in  their  solu- 
tion required  both  multiplication  and  division.  They  will  here 
be  introduced.  Let  us  take  the  following.  How  many  ells 
French  in  15  pieces  of  cloth,  containing  each  20  yards  ?  Now 
it  is  evident  that  yards  cannot  be  changed  to  ells  French,  at  a 
single  step.  When  the  question  is,  how  many  times  one  quan- 
tity is  contained  in  another,  a  simple  operation  of  division  is 
often  all  that  is  required,  to  obtain  the  answer.  But  that  will 
not  suffice  here,  and  the  reason  is,  the  two  quantities  are  not 
in  the  same  denomination.  The  first  step  then,  is,  to  bring  the 
two  quantities  given  to  the  same  kind.  The  scholar  will  there- 
fore turn  to  Table  5th,  Cloth  Measure.  He  will  there  find  that 
the  ell  French  and  the  yard,  may  both  be  brought  into  quar- 
ters ;  the  former,  by  being  multiplied  by  6,  and  the  latter,  by  4. 
Therefore, 

1  5  pieces. 

2  0— yd.  in  a  piece. 

3  0  0— yd.  in  15  pieces. 
4r=qr.  in  a  yard. 


6)120  0=rqr.  in  300  yards,  or  15  pieces. 

2  0  0  — ells  French,  the  quantity  required. 

There  are  evidently  as  many  ells  French  as  there  are  6's 
in  1200qr.  as  6  qr.  make  one  ell  French. 

The  scholar  will  perceive  that  the  given  quantity  must  be 
brought  first  into  a  denomination,  from  which  it  may  be  changed 
to  the  required  denomination. 


86  REDUCTION. 

The  above  work  may  be  abbreviated  by  applying  to  the  solu- 
tion the  principle  explained  in  the  rule  for  canceling.  The 
numbers  15,  20,  and  4,  are  multiplied  together  and  produce  the 
dividend,  1200.  This  is  then  divided  by  6,  and  the  result  is 
the  answer.  The  scholar  may  therefore  turn  back  to  the  rule 
for  canceling,  and  he  will  see  that  the  following  statement  is  in 

accordance  with  it ;  viz.  — :  .  This  statement,  by  Sec. 
5th,  of  the  same  rule,  may  be  reduced  to  '  '  - ;  and  again,  by 

Sec.  4th,  to  5'  2p '  ;  and  by  6th  and  7th  Sec.  to  200,  answer 
as  before. 

We  will  now  give  the  solution  at  a  single  statement,  thus  : 
5  2 

1S-  '   '— .  and  5x20x2=200  ells  French,  the  answer. 
ft 
% 

2.  How  many  barrels,  each  holding  2  bushels  and  3  pecks, 
are  required  to  contain  880  bushels  of  corn  ? 

The  2  bushels  and  3  pecks  equal  11  pecks.  The  question 
then  is,  how  many  times  are  1 1  pecks  contained  in  880  bush- 
els. By  the  preceding  solution,  it  will  be  seen  that  the  bush- 
els, as  they  are  to  be  divided  by  11  pecks,  must  also  be 
brought  into  pecks.  Therefore, 

880 

4  =pecks  in  one 

bushel. 

One  barrel =11  pecks,  therefore,  11)352  0=pecks  in  880 

bushels. 

320  =  number   bar- 
rels required. 

The  same  canceled.  The  scholar  will  read  over  the  explana- 
tion of  the  preceding  sum,  if  he  does  not  yet  understand  the 
work. 

Statement,  — ^r-  .     Canceled,  (see  Sec.  4,  rule  for  cancel- 

80 
ing,)        '    ,  and  80  x  4  =  320  barrels,  the  same  as  before. 

3.  In  33  guineas,  at  28  shillings  each,  how  many  pistoles, 
each  22  shillings  ? 

The  simple  question  is,  how  many  pistoles  are  there  in  33 


REDUCTION.  87 

guineas.  The  guineas  cannot  be  divided  by  the  pistoles,  for 
they  are  of  different  value.  They  must  be  brought  upon  some 
common  ground,  and  the  nearest  is  that  of  shillings  ;  therefore, 

33 

2  8 

264 
6  6 

22  s.  =  l  pistole,  therefore,  22)92  4= shillings  in  33  guineas. 

4  2= number  of   pistoles  in 
the  guineas. 

33  28 
The  same  canceled  :    -L^~.    (See  rule  for  canceling,  Sec.  4th 

3    14 
and  5th.)    Performed,  — ^--  ,  and   14x3=42,  the  number  of 

S 

pistoles,  as  before. 

4.  Purchased  24  hogsheads  of  wine,  at  1  s.  8  d  per  quart, 
and  paid  for  the  same  with  cloth,  for  which  I  was  allowed  3  s. 
4  d.  per  yard.  How  much  cloth  was  required  ? 

The  price  of  the  quart  being  given,  it  is  obvious  that  the  24 
hhd.  must  first  be  brought  to  quarts.  This  is  done  by  multi- 
plying the  24  by  63  and  by  4.  Each  quart  is  worth  Is.  8  d. 
=20  d.  Therefore,  multiplying  the  quarts  by  20  d.  gives  the 
cost  of  the  wine  in  pence.  Again,  each  yard  of  cloth  is  worth 
3  s.  4  d.  =  40  d.  If,  then,  the  pence  the  wine  cost  be  divided 
by  the  price  of  one  yard  of  cloth,  the  quotient  obtained  must 
be  the  number  of  yards  required  ;  therefore,  statement  for  can- 
celing, —  — .  Let  this  statement  be  compared  with  the 
above  analysis  of  the  sum,  and  with  the  rule  for  stating  sums 

,    ,  24.  63.  4,  2fl 
for  canceling.     Statement  repeated  and  canceled, — ; 

2 

therefore,  24x63x2=the  number  of  yards  required;  viz. 
3024  yds. 

We  will  now  give  the  usual  solution  of  this  sum,  and  observe 
the  increased  facility  of  canceling. 


88  REDUCTION, 


1  5  1  2  =  the  gallons  in  24  hhd. 
4 


6  0  4  8=qt.  in  the  same, 
2  0 


4  tO  )  1  2  0  9  6 1  Oncost  of  the  same  in  pence. 

302  4=yds.  of  cloth  required,,  the  sameas  before, 
40  d. = price  of  1  yd.  of  cloth;  therefore  I  divide  by  40. 

The  scholar  will  readily  perceive  the  object  to  be  attained  in 
sums  of  this  character  ;  viz.  to  exchange  dissimilar  quantities, 
or  quantities  of  different  denominations,  but  of  equal  value. 

RULE  FOR  THE  COMMON  MODE  OF  OPERATION. — Reduce  the 
quantity  to  be  exchanged  to  the  denomination  in  which  the 
price,  or  the  equivalent  of  exchange  of  the  other  kind,  is 
given ;  then  divide  by  this  price,  or  equivalent  of  exchange, 
and  perform  such  operations  of  reduction  as  the  nature  of  the 
case  may  require. 

RULE  FOR  CANCELING. — Consider  what  is  the  quantity  to  be 
exchanged,  and  place  it  over  a  horizontal  line  towards  the  left. 
Then  on  the  right  of  this,  also  above  the  line,  place  such  numbers 
as  are  required  to  reduce  this  quantity  to  the  denomination  in 
which  the  price,  or  equivalent  of  exchange  of  the  other  kind  is 
given.  Write  also  under  the  line  those  numbers  which  are  ne- 
cessary to  reduce  this  price,  or  equivalent  of  exchange,  to  the 
required  denomination.  Proceed  to  cancel,  multiply,  and  divide, 
as  directed  in  the  rule  for  canceling,  and  the  number  obtained 
will  be  the  one  sought. 

Note  1. — In  stating  for  canceling,  care  should  be  taken  to 
introduce  into  the  statement  every  number  required  for  the 
complete  solution  of  the  same,  including  all  operations  of 
reduction,  &c.,  since  each  number  introduced  increases  the 
opportunity  for  canceling,  and  thus  abbreviates  the  operation. 


REDUCTION.  89 

Note  2. — If  any  one  term  consists  of  more  than  one  de- 
nomination, it  should  be  reduced  to  the  lowest  denomination 
given  before  stating. 

5.  How  many  times  will  a  wheel  18  feet  in  circumference, 
turn  round  in  traveling  84  miles  ? 

The  thing  to  be  done  is  to  change  84  miles  into  revolutions 
of  the  wheel.  To  do  this,  84  miles  must  be  reduced  to  feet, 
because  the  distance  required  for  one  revolution  is  given  in  feet, 
viz.  18.  Therefore, 

84 
8 

6  7  2— the  furlongs  in  84  miles. 
4  0 

2  6  8  8  0= the  rods  in  84  miles. 
16* 


161280 
26880 
13440 

44352  O^the  feet  in  84  miles. 

Then,    18)443520(24640  Ans. 
3  6 


72 
72 

000 

The  preceding  solution  is  by  the  first  rule.     We  will  now 

solve  the  sum  by  canceling.      Statement,  -  —.      Ob- 

18 

serve  that  the  numbers  above  the  line  are  those  multiplied 
together  for  a  dividend  in  the  preceding  operation,  and  that  the 
one  below  the  line  was  the  divisor. 
8* 


90  REDUCTION, 

To  avoid  the  fraction  in  the  statement,  1  6£  may  be  written 
g-  ;  for  since  1  unit  =2  halves  or  ^,  16  units  =  32  halves  or  -^-j 

oo 

and   16^:=:33  halves  or  -~-.     The  preceding  statement  may 
therefore  be  written,  ^  Canceled,  ^l. 


Then,       4  0 
1  1 


440 

4 

1760 
1  4 

7040 
1760 

24640    The  same  Ans.  as  before. 

The  scholar  will  observe  that  the  4th  and  5th  Sec.  of  the 
rule  for  canceling,  have  been  applied  in  the  preceding  solution. 

6.  In  30  purses   containing  20   guineas   each,  how  many 
pounds  ?     Ans.  840. 

30  20    28 
Statement  for  canceling,   —L-^-  —  •     The  28  above  the  line 

expresses  the  shillings  in  a  guinea,  and  the  20  below  it,  the  shil- 
lings in  a  pound.     The  scholar  may  perform  the  solution. 

7.  How  many  pounds  in  money  will  9  tuns  of  wine  cost  at 
3s.  4  d.   per  gallon?     3s.  4  d.  =40  d.      Statement  for   can- 

9.  2.  2.  63.  40  r      _ 

£. 


For  the  terms  in   this  statement  the   scholar  is  referred  to 
Tables  1st  and  6th,  of  the  Compound  Numbers. 

8.  How  many  times  will  a  wheel  12  feet  6  inches  in  circum- 
ference, revolve  in  traveling  124  miles  ?     Ans.  52377|. 


Statement—  12  ft.  6in.  =  150in.  i  .   (See  Ta- 

150. 
ble  9th,  Compound  Numbers.) 

The  last  three  sums  have  been  stated  by  the  rule  for  cancel- 


REDUCTION.  91 

ing  only,  because  that  is  regarded  as  superior  to  the  com- 
mon mode  of  solution.  The  scholar  will  feel  at  liberty  to 
adopt  either  mode  of  operation.  The  following  sums  are  not 
stated,  that  the  scholar  may  exercise  his  own  judgment. 

9.  How  long  will  it  take  to  count  6000000,  at  the  rate  of 
75  per  minute.     Ans.  55f  days. 

10.  In  107520  pounds  of  sugar,  how  many  hogsheads,  each 
containing  6  cwt.     Ans.  160. 

11.  If  one  quart  of  melasses  cost  10  pence,  how  much  will 
12  hogsheads  cost  1     Ans.  126  £. 

12.  How  many  dollars,   each  8  s.,  will  it  cost  to  ride  45 
leagues,  at  6  pence  a  mile  ?     Ans.  $8.437-f-« 

13.  How  much  will  540  yards  of  cloth  cost  at  3  s.  4  d.  per 
yard,  in  dollars,  at  6  shillings  each  ?     Ans.  $300. 

14.  How  many  dozen  of  gallon,  quart,   and  pint  bottles,  of 
each  an  equal  number,  are  contained  in  a  cisteni  holding  144 
gallons.     Ans.  8T8r  dozen. 

15.  How  many  casks,  each  containing  1  bushel  1  peck,  are 
required  to  hold  145  bushels  ?     Ans.  116. 

16.  I  have  five  hogsheads  of  wine,  63  gallons  each,  which 
I  wish  to  put  into  gallon,  quart,  and  pint  bottles,  of  each  an 
equal  number  ;  how  many  will  be  required  ?     Anst  229,  and  1 
pint  of  wine  would  be  left. 

17.  In   16  cwt.    3  qr.  20  lb.,  how  many  parcels,  each  con- 
taining 36  Ib?     Ans.  52f. 

18.  In  56  ells  Flemish,  how  many  yards  ?     Ans.  42. 

19.  In  144  yards,  how  many  ells  French?     Ans.  96. 

20.  In  472  parcels  of  sugar,  each  72  pounds,  how  many  cwt.  ? 
Ans.  303  cwt.  1  qr.  20  lb. 

21.  If  15  casks   of  flour  contain  4000  lb.,  how  many  cwt. 
are  there  in  each?     Ans.  2  cwt.  1  qr.  14§lb. 

22.  In  81b.   of  drugs,  how  many  parcels,  each  12   drams? 
Ans.  64. 

23.  In  80  parcels,  each  15  drams,  how  many  pounds  ?  Ans. 
12i 

24.  How  many  revolutions  will  a  wheel  18  feet  4  inches  in 
circumference,  make  in  traveling  300  miles  ?     Ans.  86400. 

25.  How  many  cups,  each  weighing  22  oz.  may  be  made  of 
25  lb.  6  oz.  of  silver.     Ans.  13  cups,  and  20  oz.  silver  remain. 

26.  How  much  would  1008  nails  of  cloth  cost,  at  10  pence 
per  yard,  in  dollars,  at  6  shillings  each?     Ans.  $8.75. 

27.  In  4  bales  of  cloth,  each  15  pieces,  and  each  piece  16 
ells  English,  how  many  ells  French  ?     Ans.  800. 


92  COMPOUND  RULES. 

28.  In  6  bales,  each  12   pieces,  and   each  piece  18  yards, 
how  many  ells  Flemish  1     Ans.  1728. 

29.  In  4  ingots  of  silver,  each  weighing  2  Ib.  6oz.  11  pwt. 
how  many  grains  ?     Ans.  58656. 

30.  How  many  hours,  minutes,  and  seconds  in  one  year  ? 
Ans.  8766  hr.  525960  min.  and  31557600  sec. 

31.  In  1597  quarts,  how  many  bushels,  &c.  ?     Ans.  49  bu. 
3  pk.  and  5  qt. 

QUESTIONS. — What  are  compound  numbers?  How  do  they  increase? 
What  are  included  under  this  head  1  Let  the  14  tables  of  Compound 
Numbers  be  made  familiar,  before  the  scholar  proceeds  with  Reduction. 

What  is  Reduction  7  How  are  high  denominations  brought  into 
low  denominations  ?  And  how  are  low  denominations  brought  into 
high  1  What  is  the  rule  when  high  denominations"  are  brought 
into  low?  And  what  is  it,  when  low  denominations  are  brought  into 
high  ?  What  should  the  scholar  notice  before  commencing  to  reduce 
any  quantity  ?  In  circular  measure,  how  is  the  circle  regarded  ? 
What  are  the  divisions  of  the  circle  ?  To  what  is  this  measure  applied  ? 
How  many  days  and  hours  does  the  year  contain  ?  To  what  do  the  6 
hours  amount  in  4  years  ?  How  many  days  does  every  fourth  year 
contain]  What  is  this  fourth  year  called  ?  How  may' it  be  found  ? 
In  dividing  the  given  year  by  four,  what  does  the  figure  that  remains 
(if  any)  show  ?  What  is  the  more  usual  division  of  the  year  ?  How 
many  days  are  contained  in  each  of  the  12  months  ?  When  quanti- 
ties are  to  be  exchanged,  what  is  the  rule  for  the  common  mode  of  op- 
eration ?  What  is  the  rule  for  canceling  ?  What  is  Note  1st  ?  What 
is  Note  2d  ? 


COMPOUND    RULES. 

The  scholar  will  recollect,  that  in  simple  numbers,  the 
denominations  increase  in  value  in  the  constant  ratio  of  10. 

The  peculiarity  of  the  numbers  in  the  preceding  rule,  and  in 
the  four  next  following,  is,  that  they  have  no  uniform  ratio  of 
increase,  common  to  all  denominations ;  but  each  denomination 
has  its  own  peculiar  ratio.  These  ratios  are  all  represented  in 
the  tables  of  Compound  Numbers. 

In  simple  numbers,  10  units  make  1  ten  ;  10  tens  make  1 
hundred;  10  hundred  make  1  thousand,  &c.  In  operations 
with  these  numbers,  we  therefore  carry  for  10. 

In  the  table  of  English  money,  4  farthings  make  1  penny ; 
12  pence  1  shilling,  and  20  shillings  1  pound.  For  the  same 


COMPOUND  ADDITION.  93 

reason,  therefore,  that  we  carry  for  10  in  simple  numbers, 
we  carry  for  4, 12,  and  20,  in  operations  with  pounds,  shillings, 
pence,  and  farthings  ;  that  is,  from  farthings  to  pence,  we  carry 
for  4,  because  4  farthings  make  1  penny ;  from  pence  to  shil- 
lings by  12,  for  a  similar  reason  ;  and  from  shillings  to  pounds, 
for  20.  The  same  general  principle  and  reasoning  may  be 
applied  to  the  other  compound  tables. 

There  is  one  peculiarity  noticeable  in  writing  compound  num- 
bers. In  simple  numbers,  we  always  know  that  any  figure  sus- 
tains a  ten-fold  relation  to  the  figures  next  it ;  that  is,  the  one  on 
the  left  of  it  is  of  10  times  more  value  ;  and  the  one  on  the 
right,  of  10  times  less  value  than  they  would  be  in  its  own 
place.  Hence,  all  that  is  here  necessary,  is  that  the  figures 
preserve  their  proper  order.  In  compound  numbers,  each  de- 
nomination is  known  only  by  its  own  appropriate  mark .  There 
is,  therefore,  an  obvious  necessity  for  each  denomination  to  be 
separately  written. 


COMPOUND   ADDITION. 

Compound  Addition  is  an  operation  by  which  several  num- 
bers of  different  denominations,  as  pounds,  shillings,  pence, 
&c.  are  united  together.  The  rule  to  be  observed  in  writing 
down  these  numbers  is,  to  place  those  of  the  same  name  under 
each  other. 

Let  it  be  required  to  add  together  3  £.  15s.  9d.  3qr. ; 
5£.  6s.  8d.  2qr. ;  8  £.  13s.  lid.  3  qr. ;  and  10  £.  12s. 
8d.  2qr.  The  following  is  a  convenient  mode  of  writing 
them: 

1. 

£.      s.         d.     qr. 

315       93  The  amount  of  the  right  hand  column 

5       6       82         is    10  farthings = 2  d.    and    2qr.     Like 

8     13     11     3         simple  numbers,  the  2  qr.  are  set  down ; 

10     12       8     2         and  the  2  d.  added  to  the  column  of  pence, 

the  amount  of  which  —  38  d.  =  3  s.  and  2 

28       9       22         d.     Setting  down  the  2  d.  and  carrying 

the   3  s.  to  the   column  of  shillings,  we 

make  this  column— 19  s.=2  £.  and  9  s.     Lastly,  setting  down 


94  COMPOUND  ADDITION. 

and  carrying  as  before,  we  find  the  amount  of  the  column  of 
pounds  to  be  28,  which  we  write  at  the  foot  of  the  column. 
We  therefore  find  the  amount  of  the  four  given  numbers  to  be 
28  £.  9  s.  2  d.  2  qr. 

From  the  preceding  example,  the  scholar  will  see  the  appro- 
priateness of  the  following  rule  : 

RULE. —  Write  the  numbers  so  that  each  denomination  shall 
occupy  a  separate  column.  Then,  commencing  with  the  lowest 
denomination,  add  each  column  by  itself. 

Notice  at  the  addition  of  each  column,  to  how  many  of  the 
denomination  next  above,  the  amount  obtained  is  equal,  and  how 
many  remain.  Write  down  those  that  remain,  and  carry  the 
other  number  to  the  next  column.  Proceed  thus  through  all 
the  denominations. 

Note. — The  whole  amount  of  the  left  hand  column  must  be 
written  down,  if  it  be  in  the  highest  denomination.  If  it  be 
not  in  the  highest  denomination  it  should  be  reduced  as  far  as 
practicable. 

2. 

£.       s.        d.     qr. 

27     15       63  The  column  of  farthings  amounts  to  6 

13       7       81          qr.  — Id.     and  2  qr.        The     column    of 

24     16       90         pence  is  35  d.=2  s.  11  d.     The   column 

3       4     11     2         of  shillings  is   44s.=2£.   and  4  s.  and 

the  column  of  pounds  =  69  £.     Carrying 

69       4112         and  setting  down  agreeably  to  rule,  we 

obtain  the  annexed  amount. 
3. 

£.     s.         d.    qr. 
0     15       7     0 

0     14       63  The  farthings   are   6=ild.    2qr.     The 

0831         pence   are  28=2  s.    4  d.     The    shillings 
0     18     11     2         are  57=2  £.  17  s.  which  is  written  agree- 
ably to  the  above  note. 
2      17       4     2 

4.  5.  6. 

£.   s.    d.  qr.  £.  s.   d.  qr.  £.  s.  d. 

8  12   9  2  57  11   11  1  67  18  10 

31   6  11  0  27  13   2  3  150  19  6 

42  18   3  1  48  9   6  1  175  16  8 

2383  73  10   9  2  37  14  7 

85   1   8  2 


COMPOUND  ADDITION.  95 

7.  8.  9. 

£.        s.        d.  £.  s.  d.  £.  s.  d.    qr. 

Ill  54  7  9  444  4  11      3 

10     10     10  0  19  11  26  16  41 

100       07  144  3  10  372  0  10     2 

73       4       9  132  18  0  1780  14  6     2 

43       8     11  43  6  8  200  10  10     3 


10.  11.  12. 

£.  s.  d.  £.  s.  d.  £.  s.  d.  qr. 

76  0  0  56  18  8  875  16  10  3 

0  19  1  73  11  11  783  19  7  2 

86  11  7  22  12  2  59  17  7  3 

43  4  8  77  17  7  85  13  11  1 

750  18  6  88  18  8  387  14  9  3 


13.  Bought  a  horse  for  26  £.  12  s. ;  a  yoke  of  oxen  for  31 
£.  17s.  8d. ;  a  cow  for  7  £.   16s.  9d. ;    and  paid  15s.  8d. 
for  a  bridle.     How  much  did  they  all  cost  me  ?     Ans.  67  £. 

2  s.  1  d. 

14.  Bought  cloth  for  29  £.  6s.  10  d. ;  ribbon  for  3  s.  4  d.  3 
qr.  ;  a  pair  of  boots  for  £1.  6s.  3d.;  and  paid  2s.   8  d.  2  qs. 
for  mending  a  pair  of  shoes.     What  was  my  bill  for  the  whole  ? 
Ans.  30  £.  19s.  2  d.  1  qr. 

15.  Bought  at  one  time  goods  to  the  amount  of  175  £.  16s. 
lid.;  at  another,  to  the  amount  of  35  £.  19s.  8  d. ;  paid  for 
carting  7  £.   8s.  9  d.  3  qr. ;    and  for  insurance   3  £.   9  s.  7  d. 
What  was  the  amount  of  my  bills  1    Ans.  222  £.  14s.  11  d.  3  qr. 

16.  Sold  at  one  time  goods  to  the  amount  of  35  £    1 1  s.  6  d. 

3  qr. ;  at  another,  to  the  amount  of  56  £.  19s.  7  d.  1  qr. ;  at  a 
third  time,  to  the  amount  of  75  £.  Is.  3d.;  and  at  a  fourth,  to 
the  amount  of  63  £.  13  s.  4d.  2qr.  What  was  the  whole  amount 
of  my  sales  ?     Ans.  231  £.  5  s.  9  d.  2  qr. 

17.  Bought  a  quantity  of  corn  for   113  £.   lls.   lid.;  of 
rye,  for  32  £.  19  s.  3  d.  ;  of  wheat,  for  136  £.  16  s.  8  d.  ;  and 
of  oats,  for  22  £.  14s.   9  d.     What  was  the  whole  amount? 
Ans.  306  £.  2  s.  7  d. 

18.  A  man  sold  his   farm  for  856  £. ;  his  sheep,  for  67  £. 
17s.;  his  swine,  for  19 £.  19  s.  11  d. ;  and  his  grain,  for  36 £. 
13  s.  2  d.     How  much  money  did  he  receive  ?     Ans.  980  £. 
10s.  Id. 


96  COMPOUND  ADDITION. 


TROY  WEIGHT. 
1. 

lb.       oz.   pwt.    gr. 

15       6     13    J4  The  column  of  grains  amounts  to  69 

11  3  11  19  gr.  =  2  pwt.  21  gr.  The  pwt.  amount 
17  4  3  21  to  46=2  oz.  and  6  pwt.  The  ounces 
42  11  17  15  amount  to  26  —  2  lb.  and  2  oz.  The 
pounds  amount  to  87  ;  the  whole  of 
87  2  6  21  which  is  to  be  written  down. 

The  scholar  will  observe,  that  as  we  have  left  the  table  of 
English  money,  we  have  no  longer  to  carry  by  20,  12,  and  4. 
Our  carrying  numbers  now  are  12,  20,  and  24. 

2.  3. 

lb,        oz.  pwt.  gr.  lb.  oz.    pwt.    gr. 

]0       9     11  16  29  7      13     19 

17       9       6  8  17  6      11      11 

28     11      16  21  31  6      16     23 

36       7     17  22  71  1       18       7 


94       2  12     19            149    11  00  12 

4.                                    5.  6. 

lb.    oz.    pwt.  gr.  lb.  oz.   pwt.  gr.  lb.  oz.     pwt.   gr. 

1     8     18  12  0  9     11  16  76  11       0     21 

9     6     19  9  12  7     16  11  3333 

11     5       3  21  3  0       0  0  11  7     19       0 

22     6     10  9  71  0     16  9  13  9     11     19 

15     5     12  21  16  9     17  23  14  11      17     12 


7.  Purchased  at  one  time,  4  lb.  3  oz.  16  pwt.  15  gr.  of  silver, 
and  at  another,  7  lb.  8  oz.    18  pwt.  23  gr. ;    besides   a  quan- 
tity of  jewelry,  weighing  5  Ibs.  11  oz.  and  13  pwt.     What  was 
the  whole  weight?     Ans.  18lb.  8  pwt.  14  gr. 

8.  AddSlb.  9oz.   13  pwt.  19  gr.;  2lb.  10  oz.  9  pwt.  17  gr.; 
6  lb.  11  oz.  18  pwt.  22  gr. ;  9  oz.  11  pwt.  12  gr. ;  and  81b.  3  oz. 
6  pwt.  20grs.     Ans.  22  Ibs.  9oz.  18gr. 

9.  Again,  add  6  lb.  2  oz.  16  pwt.  14  gr. ;  3  lb.  8  pwt.  2  gr. ; 
12  lb.  4  oz.  15  pwt.  22  gr. ;  8  oz.  16  gr. ;  5  lb.  13  gr. ;  and  23  gr. 
Ans.  27  lb.  4  oz.  2  pwt.  18  gr. 


COMPOUND  ADDITION.  97 

AVOIRDUPOIS  WEIGHT. 
1. 

T.    cwt.    qr.    Ib.      oz. 

7     16     3     20     13  The  amount  of  the  ounces  is  45 

4     12     1     25     11          =2lb.  13oz.     The  pounds  amount 
3       9     2     16       9         to     77=2  qr.  21  Ib.      The   qr.    are 

12     18     0     14     12         8=2  cwt.     The  cwt.  are  57 =2  tons, 
17  cwt.  and  the  tons  are  28. 

28     17     0     21     13 

2.  3. 

T.  cwt.  qr.  Ib.  oz.  cwt.  qr.  Ib.  oz.  dr. 

3  19  0  16  15  16  3  24  15  12 

9   0  3  24  12  32  0  18  9  14 

1  18  1  26  14  7  2  21  7  9 

14   5  2  12  9  12  3  11  11  12 


4.  5. 

cwt.  qr.     Ib.  oz.  dr.  cwt.   qr.  Id.  oz.  dr. 

6     2     27  12  9  17     0  9  4  8 

12     3     16  9  12  13     3  27  15  12 

14     1     24  14  6  18     2  17  13  8 

17     3       8  15  8  29     1  23  12  13 


6.  Purchased  at  one  time,  16  tons,  3qr.  21  Ib.  of  hay;  at 
another,   9  tons,  16  cwt.  2qr.  17lb. ;  and  at  another,  27  tons, 
13 cwt.   1  qr.   17 Ib.     How  much  did  I  purchase?     Ans.  53 
tons,  10  cwt.  3qr.  27  Ib. 

7.  Bought   36  cwt.   3qr.   24 Ib.  of  wool;   but  finding  the 
demand  large,  I  made  three  successive  purchases,  at  each  of 
which  I  bought  45  cwt.  2  qr.  16  Ib.     What  was  the  amount  of 
my  purchases  ?     Ans.  173  cwt.  3  qr.  16  Ib. 

APOTHECARIES'  WEIGHT. 

1.                                   2.  3. 

ft.  ?.  3.  3.  |.  3.  3.  gr.  ft.  5.  3.  3-  gr. 

2831  6-52  12  8941   14 

5622  961  15  14  670  12 

6472  3409  19516 

8652  272  13  86725 

23  2  3  1 

9 


98  COMPOUND  ADDITION. 

4.  A  physician  purchased  the  following  quantities  of  medi- 
cine, at  three  different  times  ;  viz.  1  pound,  4  ounces,  5  drams ; 
3  pounds,  11  ounces,  6  drams,  2  scruples,  15  gr. ;  and  7  drams, 
1  scruple,  and  12  grains  ;  what  was  their  whole  weight?  Ans. 
5ib.  5|.  33.13.  7gr. 

CLOTH  MEASURE. 


1. 

2. 

3. 

4. 

yd. 

qr. 

na. 

yd. 

qr. 

na. 

E.E. 

gr. 

na. 

E.F. 

qr. 

na. 

7 

3 

2 

I 

1 

1 

16 

4 

3 

21 

3 

0 

9 

2 

3 

6 

3 

2 

12 

3 

3 

13 

5 

2 

6 

1 

0 

12 

2 

3 

18 

2 

2 

16 

4 

3 

8 

3 

3 

15 

1 

2 

20 

3 

1 

19 

2 

1 

5.  Bought  at  one  purchase,  32yd.  3qr. ;  at  another,  16yd.  2 
qr.  2  na.  ;  and  24  yd.  1  qr.  at  another.     How  many  yards  did 
I  purchase  1     Ans.  73  yd.  2  qr.  2  na. 

6.  Bought  of  one  man,  12  ells  English,  4qr.  3na. ;  and  of 
each  of  two  others,"  27  ells   English,  3  qr.   and  3  na.     How 
many  Ells  did  I  purchase  ?     Ans.  68  E.E.  2  qr.  1  na. 

7.  Received  from  France   12   ells,  4 qr.  of  broadcloth;  17 
ells,  5qr.  3na.  of  cassimere ;  and  19  ells,  2qr.  3na.  of  silks. 
How  many  ells  were  there  in  the  three  articles  purchased  ? 
Ans.  50  ells.  0  qr.  2  na. 


DRY  MEASURE. 


1. 

bu.  pJc.  qt.  pt. 
3251 
7372 

11  2  4  1 
6130 


2. 

3. 

bu.  pk.  qt.  pt. 

bu.  pk.  qt.  pt. 

8370 

15   1   6   1 

9241 

127  5  7  1 

6361 

12  2  5  0 

4220 

16  2  7  1 

WINE  MEASURE. 


1.  2.  3. 

T.  P.  hhd.gal.qt.  hhd.  gal.  qt.  pt.  gi.  hhd.  gal.  qt.  pt.  gi. 

4  1  1  42  3  15  27  3  1  2  140  46  2  1  1 

6  0  1  24  2  20  13  2  0  3  127  15  3  0  3 

8  1  0  18  3  12  16  1  1  1  263  29  1  1  2 

12  0  1  23  0  132  54  3  0  3  42  27  3  0  3 


COMPOUND  ADDITION.  99 

LONG  MEASURE. 

1.  2. 

L.  m.  fur.  rd.  m.  fur.  rd.  yd. 

12  2  5  36  9  6  12  3 

9  1  7  24  4  7  26  2 

15  0  6  17  12  4  32  5 

30  2  4  26  7  1  38  4 


3.  4. 

TO.  fur.  rd.    yd.  ft.  in.   b.  c.  L.  m.  fur.  rd.  yd.  ft.    in.  b.c. 

4     4     23     5     2  92  18  2     5  18     3     2     7     1 

9     3     30     6     1  10     1  21     0     4  30     4     1     9     2 

62     36     42  82  32     17  31     5241 

52     27     42  82  76  23  39     42   10     2 


LAND  OR  SGUJARE  MEASURE. 

1.  2. 

A.  rood.   rd.     yd.  ft.     in.  A.  rood.  rd.  yd. 

7  2     36     24  6     72  9     3     21  6 

8  3     23     20  4     91  12     2     37  11 
5     1      15     17  8  108  11     39  12 

12     3     12     13  6     22  15     3     12  16 

15     0     17       9  7  136  8     1       9  12 


3.  4. 

A.  rood.  rd.  yd.  ft.  A.   rood.  rd.  yd.  ft. 

46     0     29  11  7  34  2  33  7  6 

27     3     26  6  4  44  0  30  7  5 

18     2     32  6  4  15  3  29  10  5 

25     3     30  7  5  33  3  36  8  7 

6     1     16  6  8  44  3  37  8  7 


SOLID  MEASURE. 

1.  2.  3. 

ft.   in.  cd.  ft.   in.  cd.  ft.   in. 

99  420  11  72  726  3  99  777 

78  864  12  16  317  66  77  333 

320  740  113  17  36  122  116  1240 

950  222  4  117  1372  372  108  1617 

48   12  116  8   96  12  96  456 


100  COMPOUND  ADDITION. 


TIME. 

1.  2. 

yr.   mo.    w.    d.     h.  m.  sec.  d.  h.  m.  sec. 

2       9     3     6     13  22  56  15  21  43  51 

8     11     0     3     21  43  21  23  17  55  56 

1  12     2     4     23  69  52  6  19  59  49 

2  8     3     5     17  36  8  16  15  43  36 


3.  4. 

ID.     d.      h.  m.  sec.  w.  d.  h.  m.  sec. 

5  6       9  7  59  13  10  17  8  47 

6  5     21  39  27  14  9  20  40  43 
22     4     23  36  42  15  8  22  45  23 
11     3     16  17  18  16  11  23  18  22 


CIRCULAR  MOTION. 

1.  2.                                       3. 

S.      °        '  "  £     °        '  "  S.  ° 

6     23     42  39  49     29  41  3  22  40     37 

8     26     54  36  6     7     43  2  11  29  59     57 

3     19     11  9  11     8     51  59  4  23  18     38 

2     21      13  23  12   13     27  17  86  13     15 


APPLICATION. 


Ex.  1.  What  is  the  amount  of  23  £.  11  d.  ;   13  £.  17s.  3 
qr. ;  16s.  8  d.;   and  lid.   3  qr.  ?    Ans.  37  £.  15  s.  7d.  2qr. 

2.  Bought  the  following  quantities  of  oil ;  viz.  12  gal.  3  qt.  ; 
2  hhd.  42  gal.  2  qt.  1  pt. ;   and  13  hhd.  56  gal.      What  was  the 
whole  amount  ?    Ans.  16  hhd.  48  gal.  1  qt.  1  pt. 

3.  Add  together  250  £.  18s.  9  d.   3  qr. ;   16  £.   7s.   2  qr. ; 
21  £.  19  s.  3  d. ;   18s.  6  d.  ;  and  36  £.     Ans.  326  £.  3  s.  7  d. 
Iqr. 

4.  What  is  the  amount  of  5  cwt.   3qr.  27  Ib. ;  2  qr.  29  lb.; 
12  cwt.  1  qr.  17  lb. ;  and  36  cwt.  16  lb.  Ans.  55  cwt.  1  qr.  5  lb. 

5.  Bought  at  one  time,  7  bu.  3  pk.  of  wheat ;  at  another,  9 
bu.  1  pk.  and  had  previously  in  each  of  two  bins,  6  bu.  and  3 
pk.    What  was  the  whole  amount?    Ans.  30 bu.  2pk. 

6.  Sold  one  cow  for  10  £.  15  s.  6  d. ;  another  for  6  £.  19  s. 


COMPOUND  SUBTRACTION.  101 

lid.;  and  a  colt  for  12  j£.  6  s.  4  d.    How  much  did  they  all 
amount  to  1     Ans.  30  £.  Is.  9  d. 

7.  Bought  four  casks  of  wine,  of  which  the  first  contained 
42  gal.  2  qt.  1  pt.  ;  the  second,  65  gal.  1  pt.  ;  the  third,  50  gal. 
3  qt. ;  and  the  fourth,  55  gal.  1  qt.    1  pt.     How  many  gallons 
did  I  purchase  1     Ans.  213  gal.  3  qt.  1  pt. 

8.  Purchased  three 'pieces  of  land.     The  first  contained  17 
acres,  1  rood,  and  35  rods  ;  the  second,  36  acres,  2  roods,  21 
rods  ;  and  the  third,  46  acres  and  37  rods.     How  much  land 
did  I  purchase  ?    Ans.  100  acres,  1  rood,  13  rods. 

QUESTIONS. — In  what  ratio  do  simple  numbers  increase  1  What  is 
the  peculiarity  of  Compound  Numbers'?  Where  are  the  ratios  of  in- 
crease and  decrease  of  compound  numbers  given  1  Why  do  you  carry 
for  10  in  simple  numbers  1  Why  do  you  carry  for  4,  12,  and  20,  in 
the  table  of  English  money  1  What  peculiarity  noticeable  in  writing 
compound  numbers  1  What  only  is  necessary  in  writing  simple  num- 
bers 1  How  are  compound  numbers  known  1  How  must  each  denom- 
ination therefore  be  written  1  What  is  Compound  Addition  1  How 
are  numbers  to  be  written  1  What  is  the  rule  1  What  is  the  note 
following  the  rule  1 


COMPOUND    SUBTRACTION. 

The  scholar  has  now  become  acquainted  with  Compound 
Addition ;  and  he  was  previously  acquainted  with  the  Simple 
rules.  He  needs,  therefore,  to  be  informed  only,  that  Com- 
pound Subtraction  sustains  the  same  relation  to  Compound  Ad- 
dition, that  Simple  Subtraction  does  to  Simple  Addition.  It  is 
the  subtracting  of  numbers  of  different  denominations. 

In  this  rule,  instead  of  constantly  borrowing  10,  when  the 
lower  figure  is  the  larger,  he  must  borrow  as  many  units  as 
are  required  of  the  denomination  he  is  subtracting,  to  make  a 
unit  of  the  next  higher  denomination  ;  that  is,  when  it  be- 
comes necessary  to  borrow  a  number  in  subtracting  farthings, 
4  is  the  number  always  required  ;  in  subtracting  pence,  12  is 
the  number ;  and  in  shillings,  20  ;  and  in  like  manner  in  other 
denominations . 

RULE. — Place  the  less  quantity  under  the  greater,  so  that 
each  denomination   shall  stand  under  one  of  its  own  name  or 
kind.      Begin  at  the  right,  and  proceed  in  all  respects  as  in 
9* 


102  COMPOUND  SUBTRACTION. 

Simple  Subtraction,  except  in  borrowing  when  the  lower  figure 
is  the  larger ;  in  doing  which,  instead  of  10,  (the  number  bor- 
rowed in  Simple  Subtraction,}  borrow  as  many  units  as  make 
one  of  the  next  higher  denomination.  Whenever  a  number  is 
borrowed,  one  is  to  be  carried  to  the  next  lower  figure. 

£.       s.       d.     qr. 

Ex.  1.  From     16     18     8     1  First,  I  cannot  take  3  qr. 

Take       8     16  11     3         from   1  qr.      I  therefore 

add   4qr.  to   the    upper 

Ans.  8  182  figure,  and  take  the  3 
from  the  amount,  5,  and  obtain  a  remainder  of  2.  I  carry  1  to 
the  next  lower  figure,  viz.  11,  which  makes  it  12,  and  proceed 
to  take  it  from  the  figure  above,  but  find  it  impracticable.  I 
therefore  add  12  to  the  upper  figure,  8,  making  it  20 ;  and 
from  this  amount,  subtract  12,  and  obtain  the  8  in  the  answer. 
Again,  I  carry  1  to  the  next  figure,  16,  which  makes  it  17, 
and  take  this  from  the  figure  above,  and  obtain  a  remainder  of 
1.  I  here  borrowed  nothing  and  have  nothing  to  carry ;  there- 
fore, 8  from  16  leaves  8. 

2.  3.  4. 

£.        s.        d.  qr.  £.  s.  d.  qr.  £.  s.      d.   qr. 

35     11       9  3  46  13  7     1  74  0     9     2 

17       9     11  2  21  17  9     3  72  19  11     3 


5.  6.  7. 

£.       s.      d.  qr.  £.  s,  d.  qr.  £.  s.  d, 

99     16     8  3  33  12  3  1  94  11  8 

77     17     7  1  13  8  9  3  72       9  11 


8.  A  certain  man  owed  75  £.  13s.  9d.,  and  paid  of  this 
sum  39  £.  19s.  lid.     How  much  remained  due?     Ans.  35 
£.  13s.  lOd. 

9.  Received   of  three    individuals   the  following  sums  of 
money;    viz.  of  A.,  16  £.  12s.   8d.   3qr.;  of  B.,  21  £.   17 
s.  9d. ;  and  of  C.,  46  £.  19s.     I  afterwards  paid  D.   58  £. 
13s.  9 d.  2qr.     How  much  had  Heft?     Ans.  26 £.  15s.  8 
d.  1  qr. 

10.  The  following  sums  are  due  to  A. ;   viz.  136  £.  15  s.  11 
d.  ;  450  £.   8s.   6d.;  356  £.    17s.   10  d.  2  qr. ;  and   12  £.9 
s.  4d.     He  is  indebted  to  B.}  67  £.  14s.   9d-  2qr. ;  to  C., 


COMPOUND  SUBTRACTION.  103 

24£.  lls.   3d.;  and  to  D.,  571  £.  11s.   lid.     How  much 
is  due  to  him,  more  than  he  owes  ?     Ans.  292  £.  13s.  8  d. 

TROY  WEIGHT. 

1.  2.              3. 

Ib.  oz.  pwt.  gr.  Ib.  02.  pwt.  Ib.  oz.  pwt.  gr. 

14  9  19  16  46  11  13  9  11  11  21 

10  11  16  23  13  9  17  43  19  23 


4.                                    5.  6. 

Ib.    oz.  pwt.  gr.  Ib.  oz.   pwt.  gr.  oz.  pwt.  gr. 

36     7     14  17  89     16  11  11  9     18 

17     9     17  22  2  11     19  23  10  16     23 


AVOIRDUPOIS  WEIGHT. 

1.  2. 

cwt.   qr.      Ib.      oz.     dr.                             cwt.  qr.  Ib.      oz.      dr. 

12     3     19     13     14                            31      1  23     14     15 

9     2     21      11       6                            26     3  25     15       8 


3.  4. 

T.   &wt.    qr.     Ib.  oz.  dr.  T.  cwt.  qr.  Ib.  oz.     dr. 

6     13     0     11  12  13  9  17  3  20  15     8 

4     17     3       5  13  14  2  15  2  26  12  15 


7.  Having  in  my  possession  45  cwt.  3qr.  17lb.  of  cheese, 
Isold  32  cwt.  27  Ib.  How  much  remained?  Ans.  13  cwt. 
2qr.  18  Ib. 

APOTHECARIES'  WEIGHT. 

1.                                   2.  3. 

Ib.      I-     3.     9-    gr.      Ib.      J.    3.     3.     gr.       Ib.       |.  3-     3.    gr. 

5       9     5     2     16      12     6     5     2     15      31      17  1      1      12 

3     11     6     2     15        8     9     4     1      17      20     10  5     2     15 


CLOTH  MEASURE. 

2.                            3.  4. 

E.F.  qr.  na.             E.E.  qr.  na.  E.F.  qr.  na: 

10     5     1               21      1     2  16     1     3 

633               16     43  822 


104  COMPOUND  SUBTRACTION, 


DRY  MEASURE. 

1.                                    2.  3. 

bu.    pk.    qt.   pt.  gi.         bu.    pk.   qt.    pt.  gi.  bu.    pk.    qt.  pt.  gi. 

15     3302         26      0501  30     2713 

12     2     5     1     3         23      3     7     1     2  16     3     5     1     3* 


WINE  MEASURE. 

1.  2. 

T.    P.  hhd.  gal.    qt.   pt.  gi.  hhd.  gal.   qt.  pt.  gi. 

3     1      1     27     3     1      1  27  19     3     0     1 

2     0     2     47     1      1     3  16  43     1      1     3 


3.  4. 

hhd.     gal.    qt.  pt.  gi.  hhd.  gal.  qt.  pt.  gi. 

137     42     1  0  3  175  59  0     1     3 

128     56     3  1  1  21  37  3     1     2 


LONG  MEASURE. 

1.  2. 

m.  fur.  rd.  yd.  ft.  in.  b.c.    m.  fur.  rd.  yd.  ft.   in. 

15  4  27  4  2  11  1    32  5  39  1  2   3 

12  3  36  3  1   92    27  2  4  3  1   11 


3.  4. 

L.   m.  fur.  rd.  yd.  ft.  in.  L.  m.  fur.  rd.  yd.  ft.  in.  b.c. 

17  2  3  19  3  1  7  31  0  3  15  4  2  7  2 

12  1  7  35  4  2  8  27  2  5  17  1  2  9  2 


LAND  OR  SQ.UARE  MEASURE. 

1.                                    2.  3. 

A.    rd.     yd.  ft.  A.  rood.  rd.  yd.  A.  rood.  rd.  yd. 

9     36     14  8  74     3     27  16  12  1      16  15 

4     39       6  4  64     2     31  12  9  2     17  16 


SOLID  MEASURE. 

1.  2.  3. 

C.        ft.        in.  C.      ft.          in.  C.      ft.  in. 

21       62     856         56     110     1462  8     100  8 

16     115     972         19       36       472  1     101     1560 


COMPOUND  SUBTRACTION.  105 


TIME. 

1.  2. 

yr.  m.  w.  d.  h.      m.  «?{  d.  h.   m.   sec. 

16  8  3  5  13      12  2  1  15  21  35 

7  9  2  6  21       9  3  2  16  22  36 


3.  4. 

yr.    a.   h.  m.  sec.  yr.  d.  h.  m.  sec. 

19  152  13  42  21  .  45  67  17  50  15 

16  256  19  36  56  36  36  22  46  45 


CIRCULAR  MOTION. 


1. 

2.             3. 

s. 

o    /    // 

.  S.   o 

'   "    £.   °   ' 

" 

8 

18  42  36 

9  27 

36  51    11  15  16 

31 

6 

26  11  52 

1  29 

42  52    8  19  17 

42 

4. 

5. 

S.   ° 

/   // 

S.   °   '   " 

11  21 

49  59 

8  19  38  46 

6  27 

13  21 

6  21  42  50 

PROMISCUOUS  EXAMPLES. 

Ex.  1.  1  have  in  my  possession  46 £.  19s.  lid.  How 
much  shall  I  have  left,  after  paying  a  debt  of  27  £.  13  s.  9  d.  ? 
Ans.  19  £.  6s.  2  d. 

2.  Received  156  £.  3s.  8  d.  after  which  I  paid  out  137  £. 
15s.  lOd.     How  much  remained  in  my  possession?    Ans. 
18  j£.  7s.  10  d. 

3.  Lent  a  friend  16  £.  17s.  6d.     On  the  following  day  he 
paid  me  5  £.   13s.   lid.;    one  week  after  he  made   another 
payment  of  7 £.  5s.  lOd.      How  much  then  remained  due  ? 
Ans.  3£.  17s.  9d. 

4.  Bought  a  wagon  for  9  £.  11s.  and  sold  the  same  for  13 
£.  5  s.      How  much  did  I  gain  ?     Ans.  3  £.  14s. 

5.  Bought  a  horse  for  56  £.  15  s.  and  exchanged  the  same 


106  COMPOUND  SUBTRACTION. 

for  a  colt,  and  received  46  £.  11  s.  in  money.      How  much  did 
the  colt  cost  me  ?     Ans.  10  £.  4  s. 

6.  A  man  having  15  tons,  13  cwt.  3  qr.  of  hay,  sold  5  tons, 
10  cwt.  and  gave  3  cwt.  3  qr.  to  a  friend  ?      How  much  had  he 
left?      Ans.  10  tons. 

7.  Bought  7  cwt.  3  qr.  16  Ib.  of  rice  at  one  purchase,  and  9 
cwt.  1  qr.  27  Ib.  at  another  ;  of  this,  2  cwt.   16  Ib.  was  stolen, 
and  of  the  remainder,  I  sold  11  cwt.  2qr.  21  Ib.     How  much 
had  I  left  ?     Ans.  3  cwt.  2  qr.  6  Ib. 

8.  Three  men  bought  a  piece  of  land  for  450  £.  16  s.  10  d. 
of  which  two  of  them  paid  each  69  £.  17s.  6  d. ;  what  did  the 
third  man  pay  1    Ans.  31 1  £.  Is.  10  d. 

9.  I  owned  a  tract  of  land  containing  356  acres,  3  roods,  and 
30  rods  ;  from  this  I  sold  to  A.  127  acres,  2  roods  ;  and  to  B. 
27  acres,  1  rood,  and  36  rods.   How  much  remained  ?   Ans.  201 
acres,  3  roods,  34  rods. 

10.  A  father,  46  years,  9  months,  and  27  days  old,  has  two 
sons  ;  the  elder  of  whom  is   19  years,  3  months,  and  13  days 
old  ;  and  the  younger,  7  years,  10  months,  and  21  days.     How 
much  does  the  father's  age  exceed  the  sum  of  his  sons'  1  Ans. 
19  y.  7m.  23d. 

11.  Bought  a  quantity  of  cotton,  which,  at  the  price  agreed 
upon,  came  to  20  £.  4  s.    In  pay  for  this  I  gave  a  quantity  of 
rice  worth  15«£.  18s.  and  the  balance  in  cash.     How  much 
money  did  I  pay  ?     Ans.  4  «£.  6s. 

12.  A  merchant  bought  a  piece  of  cloth  containing  40  yards, 
from  which  he  sold  36  yd.  1  qr.  2  na.     How  much  had  he  left  ? 
Ans.  3  yd.  2  qr.  2na. 

]  3.  Sold  from  a  pile  of  wood  containing  40  cords,  64  feet, 
39  cords,  32  feet.  How  much  remained  ?  Ans.  I  cord  32  ft. 

14.  Bought  560  acres  of  land  for  940  £.  From  this  I  sold  to 
A.  120  acres,  2  roods,  and  16  rods,  for  300  £. ;  and  to  B.  150 
acres,  1  rood,  and  24  rods,  for  297  £.  10s.  and  6d.  How 
much  land  remains  in  my  possession,  and  how  much  has  it  cost 
me  1  Ans.  289  acres  ;  cost,  342  £.  9  s.  6d. 

QUESTIONS. — What  is  Compound  Subtraction  1  Instead  of  10,  how- 
many  are  to  be  borrowed  here  1  What  is  the  number  borrowed  in  sub- 
tracting farthings'?  Why  1  What  in  subtracting  pence  1  Shillings'? 
And  why?  What  is  the  Rule? 


COMPOUND  MULTIPLICATION.  107 


COMPOUND    MULTIPLICATION. 

The  peculiarity  of  Compound  Numbers  having  been  fully 
explained,  and  multiplication  of  Simple  Numbers  being  also 
understood,  no  other  definition  of  this  rule  is  needed  than  is 
conveyed  by  the  name. 

A  simple  inquiry  presents  itself,  viz.  are  both  the  given  num- 
bers compound  ?  To  answer  this,  the  scholar  needs  only  to  be 
informed,  that  in  multiplication,  the  multiplier  is  always  a  sim- 
ple number,  showing  how  many  times  the  multiplicand  is  to  be 
repeated.  The  multiplicand,  therefore,  only  is  compound. 
The  product,  as  it  is  formed  by  repeating  the  multiplicand, 
must  necessarily  be  of  the  same  denomination  with  it. 

Take  the  following  illustration.    Multiply  8  s.  9  d.  2  qr.  by  4. 

PERFORMED. 

8s.     9  d.     2  qr. 
4 


1£.    15s.    2d.     Oqr. 

In  this  example,  we  say,  four  times  2  farthings  are  8  far- 
things, and  8qr.=2d.  and  Oqr.  remain.  We  therefore  write 
down  a  cypher,  and  carry  2.  Again,  four  times  9  d.  are  36  d. 
and  2  d.  to  carry  make  38  d.  =  3  s.  and  2  d.  The  2  d.  is  writ- 
ten down,  and  the  3  s.  carried.  Lastly,  four  times  8  s.  are  32  s. 
and  3  s.  to  carry  make  35  s.  =  1  £.  15s.  which  is  written  down 
as  seen  in  the  answer.  If  now,  in  the  above  example,  8s.  9 
d.  2qr.  had  been  given  as  the  price  of  one  yard  of  cloth,  and 
the  scholar  had  been  required  to  find  the  price  of  4  yards,  the 
operation  would  have  been  the  same.  The  number  of  yards 
only  decides  how  many  times  the  price  of  one  yard  is  to  be  re- 
peated. 

CASE  1st. — WHEN    THE  MULTIPLIER  OR    SIMPLE   NUMBER  is 

NOT  GREATER  THAN   12. 

RULE. — Multiply  the  Compound  number  by  the  Simple  one, 
commencing  with  the  lowest  denomination  and  carrying  from 
one  denomination  to  another,  as  in  the  preceding  Compound 
Rules. 


108  COMPOUND  MULTIPLICATION. 

Ex.  1.  Multiply  9  £.  16  s.  8  d.  2  qr.  by  9. 

PERFORMED. 

9£.     16s.     8d.     2.qr. 
9 


88  £.   10s.      4d.     2qr.  Ans. 

Explanation. — 9  times  2  qr.  =  18  qr.  —  4  d.  2  qr.  9  times  8  d. 
=72  d.  and  4  d.  from  the  farthings  added,  make  76  d=6  s.  4  d. 
Again,  9  times  16s.  =  144s.  and  the  6s.  obtained  from  the 
pence  make  150  s.  =7  £.  10  s.;  and  lastly,  9  times  9  pounds^ 
81  £.  and  81  £.  +  7  £.  =  88.  The  product  therefore  is,  as 
given  above,  viz.  88  £.  10s.  4d.  2qr. 

2.  Multiply  16  £.  11  s.  9  d.  3qr.  by  3. 

PERFORMED. 

16  £.     11s.     9d.     3qr. 
3 


49  £.     15s.     5d.     Iqr.  Ans. 

3.  Multiply  1  £.   11s.  6  d.  2  qr.  by  5.     Ans.  7  £.  17  s.  8 
d.  2  qr. 

4.  Multiply  11  s.  9  d.  by  3.     Ans.  1  £.  15  s.  3  d. 

5.  Multiply  15  £.  10  s.  8  d.  by  2.     Ans.    31  £.  I  s.  4  d. 

6.  Multiply  5  s.  6  d.  by  9.     Ans.  2  £.  9  s.  6  d. 

7.  What  cost  4  gallons  of  wine  at  8  s.  7  d.  per  gallon?  Ans. 
l£.  14s.  4d. 

8.  What   cost   5  cwt.   of  raisins  at  1  £.  7s.  9d.  2  qr.  per 
cwt.?     Ans.  6£.  18s.  lid.  2qr. 

9.  What  cost  8  yards  of  broadcloth  at  1  £.  2  s.  3d.  per  yard? 
Ans.  8£.  18s. 

10.  What  cost  11    tons  of  hay  at  2  £.  1  s.  10  d.  per  ton? 
Ans.  23  £.  0  s.  2  d. 

11.  What  cost  12  bushels  of  wheat  at  9  s.  10  d.  per  bushel  ? 
Ans.  5£.  18s. 

CASE  2d. — WHEN  THE  MULTIPLIER  OR  SIMPLE  NUMBER  is  A 

COMPOSITE  NUMBER  GREATER  THAN  12. 

RULE. — Separate  the  simple  number  or  multiplier  into  its 
component  parts,  and  multiply  first  by  one  of  these  parts,  and 
the  product  of  this  multiplication  by  the  others  in  succession. 
The  last  product  will  be  the  answer  required. 


COMPOUND  MULTIPLICATION.  109 

Note. — It  will  generally  be  found  more  expeditious  to  divide 
the  multiplier  into  two  parts  only.  Should  it,  however,  be  large, 
as  125  or  1728,  it  may  be  divided  into  more  than  two  parts, 
viz.  125  into  three  5's,  and  1728  into  three  12's. 

Ex.  1.  Multiply  6s.   10  d.  by  28.     28  =  4x7.     Therefore, 

6s.     10  d. 

.  7 

2  £.     7s.     1 0  d.  = product  of  7. 
4 


9£.  11  s.      4  d.= prod,  of  4  times  7 =28. 

2.  What  cost  27  yards  of  cloth  at  7  s.  6  d.  per  yard  ?     27  = 
9x3.     Therefore, 

7s.     6  d. 
9 


3  £.  7  s.     6  d.=price  of  9  yards. 
3 


10.£.  2s.      6d.=priceof  3  times  9yd.  =  27yd. 

3.  What  cost  32  yards  at  9  s.  9  d.  per  yard  ?      Ans.  15  £. 
12s. 

4.  What  cost  20  yards  at  3s.  6  d.  per  yard  ?    Ans.  3  £.  10s. 

5.  What  cost  36  gallons  at  5  s.  8  d.  per  gallon  ?     Ans.  10  £. 
4s. 

6.  What  cost  63  yards  at  7  s.  6  d.  2  qr.  per  yard  ?     Ans.  23 
J6.  15s.  Id.  2qr. 

7.  What  cost  72  yards  at  3  s.  lid.  per  yard  ?      Ans.    14£. 
2s. 

8.  What  cost   ]44  yards   at  1  £.  4  s.  2  d.  per  yard?     Ans. 
174  £,. 

9.  Sold  21  men  each  3  qr.  16  Ib.  of  sugar.    How  many  cwt. 
did  I  sell  ?     Ans.  18  cwt.  3  qr. 

10.  Suppose  27  young  lads  to  hare  lived  each  6  years,  9 
months,  8  days,  and  11  hours  ;  how  many  days,  <fcc.,  have  they 
all  lived,  allowing  30  days  to  a  month  ?     Ans.  182  yr.  10  mo. 
18  d.  9h. 

10 


110  COMPOUND  MULTIPLICATION. 


CASE  3d.  —  WHEN  THE  MULTIPLIER  is  MORE  THAN  12,  AND  is 

NOT  A  COMPOSITE  NUMBER. 

RULE.  —  Take  any  two  or  more  numbers  whose  product  will 
come  as  near  as  possible  to  the  given  number  or  multiplier, 
without  exceeding  it,  and  having  multiplied  the  given  price  of 
one  yard,  pound,  <fyc.  by  them,  retain  their  product.  Then 
multiply  the  same  given  price  by  the  riumber  wanting  to  make 
up  the  entire  multiplier,  and  add  the  product  to  the  preceding 
product.  Their  sum  will  be  the  product  required. 


.  —  If  preferred,  the  given  quantity  or  multiplier  may  be 
multiplied  by  each  denomination  of  the  compound  quantity 
separately,  and  the  several  products,  reduced  to  the  highest 
denomination,  may  be  added  together. 

Ex.  1.  What  will  51  yards  of  cloth  cost  at  3  s.  6  d.  per 
yard?  The  two  numbers  whose  product  comes  nearest  to  51, 
are  7  and  7,  and  their  product  is  49.  Consequently,  if  we 
multiply  3  s.  6  d.  by  7,  and  their  product  by  7  again,  we  shall 
obtain  the  cost  of  49  yards.  There  will  then  be  the  cost  of 
two  yards  wanting.  This  will  be  obtained  by  multiplying  3  s. 
6  d.  the  price  of  1  yard,  by  2.  The  operation  is  thus  performed  : 


3s. 

6d. 

7 

1  £. 

4s. 

6d.=the  price  of  7  yards. 

7 

8£. 

11s. 

7s. 

6d.=price  of  49  yards. 
Od.:=  price  of  2  yards. 

8£. 

18s. 

6d.=rprice  of  51  yards. 

Or,  the  sum  may  be  solved  by  the  note,  thus  :  51  yards,  at 
6  d.  per  yard,  would=3C6  d.^1  £.  5  s.  6  d.  ;  and  51  yards  at  3 
s.  per  yard=153  s.  =  7  £>  13  s.  ;  and  1  £.  5  s.  6  d.  added  to  7 
£.  13s.  as  before,  gives  8  £.  18  s.  6  d. 

2.  What  cost  47  yards  of  cloth,  at  17  s.  9  d.  per  yard  ?  The 
two  numbers  required  by  the  rule  are  6  and  7,  and  there  is  a 
remainder  of  5.  Therefore, 


COMPOUND  MULTIPLICATION.  Ill 

17s.     9d. 

6 


5£.       6s.     6  d.= price  of  6  yards. 

7 


37  £.       5s.     6d.=price  of  42  yards. 
4£.       8s.     9d.=price  of  5  yards  added. 


41  £.     14s,     3 d.=  price  of  47  yards. 

Or,  47yds.  x    9  d.=   1  £.  15s.  3d. 
47yds  x!7s.  =  39£.  19s.  0  d. 

41  £.   14s.  3d.  as  before. 

3.  What  cost  23  gallons  of  melasses  at  3  s.  6  d.  per  gallon  ? 
Ans.  4  £.  6  d. 

4.  What  cost  94  yards  of  cloth  at  1  £.  9  s.  4  d.  per  yard  ? 
Ans.  137  £.  17s.   4  d. 

5.  What  cost  59  yards  of  baize  at  3  s.  4  d.  per  yard  ?     A?is. 
9£.  16s.  8d. 

6.  What  cost  29  cwt.  of  sugar  at  17s.  8d.  per  cwt.  ?     Ans. 
25  £.  12s.  4d. 

7.  What  cost  78  yards  of  cloth  at  9  s.  3  d.  per  yard  ?     Ans. 
36  £.  Is.  6d. 

8.  What  cost  65  cwt.  of  sugar  at  19  s.  3  d.  per  cwt.  ?     Ans. 
62  £.  11s.  3d. 

9.  Seventeen  men  brought  each  a  load  of  hay  to  market, 
weighing  17  cwt.  3  qr.  and  21  Ib.  and  received  each  for  his 
load,  5  £.  8s.  3d.     What  quantity  of  hay  did  they  all  bring ; 
and  how  much  money  did  they  all  receive  1    Ans.  They  brought 
15  T.  4  cwt.  3  qr.  21  Ib.  and  received  92  £.  0  s.  3  d. 

EXAMPLES  OF  WEIGHTS  AND  MEASURES. 

1.  What  is  the   weight  of   5    hogsheads   of  sugar,    each 
weighing  7  cwt.  3  qr.  16  Ib.  1     Ans.  39  cwt.  1  qr.  24  Ib. 

2.  What  is  the  weight  of  9  chests   of  tea,  each  weighing 
3  cwt.  2  qr.  9  Ib.  ?     Ans.  32  cwt.  25  Ib. 

3.  In  8  piles  of  wood,  each  containing  4  cords  and  56  feet, 
how  many  cords  and  feet  ?     Ans.  35  cords,  64  feet. 

4.  Multiply  15  yards,  3  qr.  and  2  nails,  by  9.     Ans.  142  yd. 
3  qr.  2  nails. 

5.  Multiply  20  years,  5  months,  3  weeks,  and  6  days,  by  14. 
Ans.  286  yr.  U  mo.  3  w, 


112  COMPOUND  DIVISION. 

6.  In  10  fields,  containing  each   12  acres,  2  roods,  and   16 
rods,  how  many  acres  ?     Ans.  126. 

7.  In  7  casks,  containing  each  42  gallons,   3  quarts,   and  1 
pint,  how  many  gallons,  &c  ?     Ans.  300  gal.  1  pint. 

QUESTIONS. — What  is  always  the  nature  of  the  multiplier!  Which 
is  the  compound  number,  the  multiplier  or  multiplicand?  What  is  the 
nature  of  the  product  1  In  case  the  quantity  by  which  you  multiply  is 
yards,  what  does  the  number  of  yards  decide  1  What  is  Case  1st  7 
"What  is  the  rule  1  What  is  Case  2d  1  What  is  the  rule  1  What  is 
the  note  under  Case  2d  1  What  is  Case  3d  1  What  is  the  rule"? 
What  note  follows  the  rule  1  No  direct  definition  has  been  given  of 
Compound  Multiplication  ;  will  the  scholar  give  one  1 


COMPOUND    DIVISION. 

This  is  the  last  of  the  Compound  Rules,  and  is  the  reverse 
of  the  preceding.  In  this  rule  a  compound  number  is  given  as 
a  dividend,  and  a  simple  number  as  a  divisor ;  and  by  the  opera- 
tion the  dividend  is  resolved  into  as  many  equal  parts  as  there 
are  units  in  the  divisor.  The  quotient  is  always  one  of  these 
equal  parts,  and  is  therefore  a  compound  number  ;  each  figure  of 
which  is  of  the  same  denomination  as  the  figure  or  figures  in 
the  dividend  from  which  it  was  obtained. 

CASE    1st. — WHEN  THE  DIVISOR  OR  SIMPLE  NUMBER  is  12,  OR 

LESS  THAN   12. 

RULE. — Divide  the  highest  denomination  first.  If  after  divi- 
ding this,  there  be  a  remainder,  reduce  it  to  the  next  lower 
denomination,  adding  the  figures  of  the  dividend  in  that  denomi- 
nation to  it,  and  divide  again.  Proceed  in  the  same  manner 
through  all  denominations ;  the  number  obtained  will  be  the  one 
required. 

Ex.1.  Divide  17  £.  11s.  5  d.  by  8. 

PERFORMED. 
8  )   17£.      US.  5d, 


345.      3s, 


COMPOUND  DIVISION.  113 

17  .£.  4-8=2  and  1  remains.  What  remains  of  any  number 
or  quantity,  must  obviously  be  of  the  same  kind  as  the  quantity 
itself.  Therefore,  the  1  is  one  pound=20s.  and  20s. +  11  s. 
=  31  s.  and  3 1-4- 8  =  3s.  and  7  s.  remain.  Again,  7s.  =  84d. 
and84d.-|-5d.  =  89d.;  and  89 ^8  =  11  and  1  remainder  =  111 d. 
Therefore  the  answer  is  as  given  above,  viz.  2  £>.  3  s.  Ill  d. 

2.  Divide  25  £.  18  s.  9  d.  by  6. 

PERFORMED. 

6)25£.     18s.     9d. 


4£.       6s.     5d.  2qr. 

3.  Divide  13  £.  19  s.  4  d.  by  4.     Ans.  3  £.  9  s.  10  d. 

4.  Divide  140 £.  12s.  9d.  by  12.     Ans.  11  £.  14s.  4d.  3qr. 

5.  Divide  73  £.  16s.  lid.  3  qr.  by  9.     Ans.   8£.  4  s.  1  d. 
Ifqr. 

6.  Divide  12  cwt.  3  qr.  12  Ib.  by  10.     Ans.  1  cwt.  1  qr.  4lb. 

7.  Eleven  men  own   equal  shares  of  36  hhd.  42  gal.  and  2 
qt.  of  wine  ;   what  is  each  man's  share  ?     Ans.  3  hhd.  21  gal. 
Oqt.  Opt.  1/T  gills. 

8.  Seven  men  bought   16  hhd.  24  gal.   3  qt.  of   wine,  for 
which  they  paid  45  £.   18  s.  6  d. ;  each  man  paying  the   same 
money  and   consequently  entitled  to  an  equal   share   of  wine. 
What  was  each  man's  share,  and  how  much  money  did  he  pay? 
Ans.   His  share  was  2  hhd.   21  gal.  2|  qt.   and  paid  6  £.    Is. 
2  d.  24  qr. 

CASE  2d. — WHEN    THE    DIVISOR   is    A  COMPOSITE  NUMBER 

GREATER    THAN     12. 

RULE. — Resolve  the  divisor  into  its  component  parts,  and 
divide  the  compound  number  by  each  of  these  parts  in  succession. 
The  quotient  arising  from  the  first  division  will  form  a  divi- 
dend for  the  second ;  and  so  on. 

Ex.  1.  Divide  26  £.  16  s.  8  d.  by  21 .  The  factors  of  21  are 
7  and  3,  because  7x3=21.  Therefore, 

7)26£.      16s.     8d. 
3)3£.      16s.     8  d. 


1£.       5s.     6fd.=the  quotient   of    26  £.    16s. 
8d.-f-21,  and  is  the  answer. 
10* 


114  COMPOUND  DIVISION. 

2.  Divide  47  £.  15  s.  8  d.  by  24.     Ans.  1  £.  19  s.  9  d.  3£qr. 

3.  Divide  85  £.  11  s.  lid.  by  16.  Ans.  5  £.6  s.  11  d.  s|qr. 

4.  Divide  128  £.  9  s.  by  42.     ^w*.  3  £.  1  s.  2  d. 

5.  Divide  15  £.  18  s.  9  d.  by  72.     Ans.  4  s.  5&  d. 

6.  Divide  5£.  10s.  3d.  by  81.     Ans.  1  s.  4d.  l£qr. 

7.  Divide  7  £.  19  s.  9  d.  by  96.     Ans. 1  s.  7  d.  3£qr. 

8.  Divide  27  £.  16  s.  by  32.     Ans.  17  s.  4  d.  2  qr. 

CASE  3d. — WHEN  THE  DIVISOR  is  LARGE  AND  NOT  A  COMPOSITE 
NUMBER. 

RULE. — Divide  the  whole  compound  quantity  by  the  whole 
divisor ;  reducing  the  remainders  after  the  division  of  each 
denomination  to  the  next  lower  denomination,  as  directed  in 
Case  1st. 

Ex.  1.  Divide  8  £.  18s.  6  d.  by  51. 

PERFORMED. 

51  )  8JE.     18s.     6  d.  (  OJ6.     3s.     6  d. 
20 

178= the  shillings  in  8  £.  18s. 
153 

25= shillings  remaining. 
12 

306=pence  in  25  s.  6  d. 
306 

000 

The  pounds  are  first  reduced  to  shillings,  and  the  given  shil- 
lings are  added.  The  178  shillings  are  thus  produced.  This 
divided  by  51  gives  3  as  a  quotient  figure  and  25  as  a  remain- 
der. After  a  second  reduction,  306  pence  are  obtained,  which 
contains  the  divisor  six  times.  Thus  the  answer  obtained  is 
3s.  6  d. 

2.  Divide  41  £.  14  s.  3  d.  by  47.     Ans.  17  s.  9  d. 

3.  Divide  4  £.  I  s.  5  d.  2  qr.  by  23.    Ans.  3  s.  6  d.  2  qr. 

4.  Divide  137  £.  17  s.  4  d.  by  94.    Ans.  I  £.  9  s.  4  d. 

5.  Divide  36 £.  1  s.  6 d.  by  78.    Ans.  9s.  3d. 

6.  Divide  10  £.  5  s.  8  d.  by  59.    Ans.  3  s.  5  d.  3  qr.  -f- 

7.  Divide  25  £.  12s.  4  d.  by  29.    Ans.  17s.  8  d. 

8.  Divide  61  £.  12  s.  by  65.    Ans.  18  £.  1 1  s.  1  qr.+ 


COMPOUND   DIVISION.  115 


EXAMPLES  IN  WEIGHTS  AND  MEASURES. 

1.  Divide   5hhd.  42  gal.  3qt.   equally  among  4  men.     Ans. 

1  hhd.  26  gal.  1  qt.  1  pt.  2  gi. 

2.  Divide  14cwt.  1  qr.  12lb.by5.    Ans.  2  cwt.  3  qr.  13  Ib.  9 
oz.  9f  dr. 

3.  Divide  27yd.  1  qr.  2na.  by  7.    Ans.  3yd.  3qr.  2-f-  na. 

4.  Divide  156  bushels,  3  pk.  6qt.  by  18.    Ans.  8bu.  2pk.  7qt. 

5.  Divide  9  hhd.  28  gal.  2  qt.  by  12.    Ans.  49  gal.  2  qt.  1  pt. 

6.  Divide  16cwt.  3qr.  18  Ib.  by  32.    Ans.  2qr.  3  Ib.  3  oz. 

7.  If  27  loads  of  hay  weigh  30  tons,  8  cwt.  2  qr.  23  Ib.  what 
is  the  weight  of  one  load  ?     Ans.  1  ton,  2  cwt.  2  qr.  5  Ib. 

8.  A  man  traveled   17  leagues,   1  mile,  4  furlongs,  and  21 
poles,  in  21  hours.     At  what  rate  did  he  travel  per  hour?     Ans. 

2  m.  4  fur.  and  1  pole. 

9.  Nine  men  own  56  Ib.  6oz.  and  17pwt.  of  silver.     What 
will  each  man  receive  if  the  whole  quantity  be  equally  divided 
among  them?     Ans.  61b.  3  oz.  8pwt.  13£gr. 

10.  Bought  15  loads  of  hay,  the  whole  weight  of  which  was 
12  tons,  15  cwt.  3  qr.  161b.     Supposing  them  all  to  have  been 
equal,  what  was  the  weight  of  each  ?     Ans.  17  cwt.  6f  Ib. 

11.  If  a  man's  income  be86«£    18s.  10  d.  per  year,  what  is 
it  per  calendar  month  ?    Ans.  7  £.  4  s.  lOf  d. 

12.  If  I  pay   15  £.  3s.  8  d.  for  56  pairs  of  gloves,  what  is 
one  pair  worth  ?    Ans.  5  s.  5  d   Of  qr. 

13.  If  a  hogshead  of  wine  cost  33  £.    12s.,  what  is  the 
price  of  a  gallon  ?     Ans.  10s.  8  d. 

14.  If  42  yards  of  cloth  cost  21  «£.  18  s.  8  d.,  what  was  the 
cost  per  yard  ?    Ans.  10s.  5  d.  H  qr. 

15.  If  16  men  cut  53  cords,  69  feet  of  wood  in  2  days,  what 
did  each  man  cut  per  day  ?     Ans.  1  cord,  86^5¥  feet. 


APPLICATION  OF  THE  FOUR  PRECEDING  RULES. 

1  .  A  silversmith  sold  to  his  customer  3  dozen  silver  spoons, 
each  weighing  3oz.  3pwt.  16  gr.  ;  l£  dozen  tea  spoons,  each 
weighing  14  pwt.  20  gr.  ;  3  silver  cups,  each  weighing  20  oz. 
18  pwt.  In  return,  he  received  old  silver  to  the  amount  of  8  Ib. 
1  1  oz.  19  pwt.  ;  for  how  much  ought  he  to  receive  pay  ?  Ans. 
61b.  lOoz.  14  pwt. 

2.  Bought  the  following  articles  at  the  prices  mentioned;  viz. 


116  COMPOUND  DIVISION. 

£.  S.  d. 

4  cwt.  of  sugar  at  2  £.  4s.  8  d.  per  cwt. 

3  hhd.  of  melasses  at  2  s.  4  d.  per  gallon, 

5  Ib.  of  green  tea  at  7  s.  6  d.  per  pound, 
12  Ib.  of  raisins  at  2  s.  per  pound, 

42  yards  of  cotton  cloth  at  Is.  6  d.  per  yard, 
27  pounds  of  ham  at  1  s.  3d.  per  pound  ; 

What  was  the  amount  of  my  bill  ?     Ans.38£.     17s.     lid. 

3.  Bought  of  James  Rankin,  £.         s.         d. 
27  yards  of  broadcloth  at  1  £.  9s.  per  yard, 

42  yards  of  Irish  linen  at  4  s.  6  d.  per  yard, 

36  hats  valued  each  at  18s. 

30  pairs  of  shoes  at  7  s.  8  d.  per  pair  ; 

What  was  the  amount  of  my  bill  ?    Ans.  92  £.     10s.     0  d. 

4.  A.  owning  100  acres  of  land,  divided  it  into  8  equal  parts, 
and  sold  each  part  for   $22.50  per  acre.     How   many  acres 
were  there  in  each  part,  and  what  was  the  value  of  the  same  ? 
Ans.    12  A.  2R.  ;  value,  $281.25. 

5.  Out  of  a  pipe  of  wine  a  merchant  sold  36  gallons,  3  quarts, 
and  1   pint  at  each  of  three  different  times;  he   then  rilled  15 
bottles,  holding  1  pint  and  2  gills  each  ;    how  much  remained  ? 
Ans.  12  gal.  2  qt.  and  2  gills. 

6.  Bought   144  pairs  of  shoes  for  96  £. ;    what  was  the 
price  of  one  pair?     Ans.  13  s.  4d. 

7.  A  person  dying,  left  real  estate  to  the  amount  of  2356  £. 
19  s.  9  d.  and  personal  property  to  the  amount  of  3184  £.  12  s. 
8  d.     In  his  will,  he  directed  that  his  wife  should  receive  one 
sixth  of  the  whole,  and  that  the  remainder  should  be  equally 
divided  among  his   four  daughters.     What  was  the   share   of 
each  ?     Ans.  The  widow,  923  £.  12  s.  |  d.  and  the  daughters 
each  1 154  £.  10s.  Ijyd. 

QUESTIONS. — How  does  Compound  Division  compare  with  Com- 
pound Multiplication  1  What  is  given  as  a  dividend  1  What  as  a 
divisor  1  Jnto  what  is  the  dividend  resolved  by  the  operation  1  What 
is  always  one  of  these  equal  parts  1  Is  the  quotient  a  Simple  or  Com- 
pound Number  1  How  may  you  know  the  denomination  of  each 
figure  in  the  quotient  1  What  is  Case  1st  1  What  is  the  rule  1  What 
is  Case  2d  'I  What  is  the  rule  1  What  is  Case  3d  ?  What  rule  1 


VULGAR  FRACTIONS. 


VULGAR   FRACTIONS. 

1 .  When  a  unit  or  single  object  is  divided  into  a  number  of 
equal  parts,  each  of  these  parts  is  a  fraction. 

If  it  be  divided  into  two  equal  parts,  each  part  is  called  a 
half,  and  is  thus  written,  ^. 

If  it  be  divided  into  three  equal  parts,  each  is  called  a  third, 
and  is  thus  written,  £. 

If  the  Avhole  be  separated  into  six  equal  parts,  each  part  is 
called  a  sixth,  and  if  into  eight  equal  parts,  an  eighth,  of  the 
whole,  and  are  thus  written,  1, }. 

When  more  parts  than  one  are  to  be  expressed,  the  figure 
above  the  line   designates  their  number,  thus,  %  ;  by  which 
expression,  we  are  to  understand  that  the  unit  is  divided  into|i| 
six  equal  parts,  and  that  five  of  these  parts  are  included  in  the" 
fraction. 

The  fraction  therefore  is  used  to  express  parts  of  units,  and 
is  represented  by  two  numbers,  one  standing  below  and  the 
other  above  a  short  horizontal  line.  The  number  below  the 
line  is  called  the  denominator,  and  shows  the  number  of  equal 
parts  into  which  the  unit  is  divided.  The  number  above  the 
line  is  called  the  numerator,  and  shows  how  many  of  these 
equal  parts  are  included  in  the  fraction,  or  make  up  its  value. 
Thus  of  the  fraction  | ,  the  lower  number  shows  a  unit  to  be 
divided  into  nine  equal  parts  ;  and  the  upper  number,  that  five 
of  these  parts  are  included  in  the  fraction. 

These  two  numbers,  when  spoken  of  collectively,  are  called 
the  terms  of  the  fraction. 

2.  Fractions   are   divided  into  six  kinds  ;  viz.  Proper,  Im- 
proper, Simple,  Compound,  Mixed,  and  Complex. 

A  Proper  Fraction  is  one  whose  numerator  is  less  than  its 
denominator,  as,  f . 

An  Improper  Fraction  is  one  whose  numerator  equals  or 
exceeds  its  denominator,  as,  J. 

A  Simple  Fraction  consists  of  one  expression,  and  is  either 
proper  or  improper,  as,  f  or  £. 

A  Compound  Fraction  is  the  fraction  of  a  fraction,  as,  |  of  J 
of  £.  It  may  consist  of  any  number  of  simple  fractions. 

A  Mixed  Number  consists  of  a  whole  number  and  fraction, 
written  together,  as,  6|,  251,  &c? 


118  VULGAR  FRACTIONS. 

A  Complex  Fraction  is  one  that  has  a  fraction  in  its  numerator 
or  denominator,  or  both,  as,  ~,  ^,  ||,  &c. 

3.  The  denominator  shows  the  number  of  equal  parts  into 
which  the  unit  is  divided  ;  and  the  numerator,  how  many  of 
these  parts  are  expressed  in  the  fraction.  Consequently,  the 
greater  the  numerator,  the  denominator  being  given,  the  greater 
the  value  of  the  fraction  ;  and  the  less  the  numerator,  the  less 
the  value  of  the  fraction.  If  the  denominator  be  8,  and  the 
numerator  1 ,  the  value  expressed  is  J,  or  one  eighth  part  of  a 
unit ;  if  the  numerator  be  2,  the  value  expressed  is  f,  or  two 
eighth  parts  of  a  unit ;  if  it  be  4,  the  value  is  -|,  or  four  eighth 
parts  of  a  unit ;  and  if  it  be  6,  the  value  is  £,  or  six  eighth  parts 
of  a  unit. 

The  value  of  a  fraction  is  therefore  the  quotient  arising  from 
t .  dividing  the  numerator  by  the  denominator,  and  always  increases 
in  the  same  ratio  as  the  numerator,  so  long  as  the  denominator 
remains  unaltered. 

We  may  therefore  express  any  value,  not  only  less  than  a 
unit,  but  equal  to  and  even  greater  than  a  unit,  by  a  fraction. 
Thus,  if  we  take  9  as  the  denominator  of  a  fraction,  and  any 
number  less  than  9  as  a  numerator  of  the  same,  the  value  ex- 
pressed is  always  less  than  a  unit,  as,  f- ;  or  if  9  be  taken  as  the 
numerator,  we  obtain  the  fraction  f ,  which,  as  the  unit  was 
divided  into  9  parts  only,  is  obviously  equal  to  1.  Again,  we 
may  suppose  more  than  a  single  unit  of  the  same  kind  to  be 
divided  in  the  same  manner,  and  their  parts  united  in  one 
fraction,  and  thus  obtain  fractions  of  any  value  more  than  a  unit. 
If  two  units  be  thus  divided  into  seven  equal  parts,  and  three 
parts  of  the  one  be  united  to  all  the  parts  of  the  other,  the 
fraction  would  be  ^  ;  or  if  all  the  parts  of  each  be  united,  it 
would  be  y,  which  is  equal  to  2  ;  or  if  three  units  were  thus 
divided,  all  their  parts  would  produce  the  fraction  2y=3. 

The  only  consideration  which  limits  the  value  01  a  fraction, 
is  the  number  of  equal  parts  united  in  the  same  expression. 

From  the  preceding  it  is  obvious  that  the  value  of  a  fraction 
is  increased  in  the  same  ratio  as  the  numerator  ;  hence, 


4.  A  fraction  is  multiplied  by  a  whole  number,  by  multiplying 
the  numerator  only. 


VULGAR  FRACTIONS.  119 

In  accordance  with  the  above  principle,  the  scholar  may 
multiply  the  following  examples  : 

1.  Multiply  £  by  3.     Ans.  f  . 

2.  Multiply  |  by  5.     Ans.  \°. 

3.  Multiply  |  by  3.      Ans.  f  . 

4.  Multiply  1  by  9.     Ans.  f  . 

5.  Multiply  T8?  by  6.     Ans.  f|. 

6.  Multiply  ||  by  9.     Ans.  f|. 

7.  Multiply  ^  by  7. 

8.  Multiplyf  by  12. 

9.  Multiply  iiby  12. 

10.  Multiply  I  by  8. 

11.  Multiply  $•  by  3.      An*.  J=L 

12.  Multiply  i  by  7.     An*.  |=1. 

From  the  last  two  examples,  it  is  obvious  that  a  fraction  is 
multiplied  by  a  number  equal  to  its  own  denominator,  by 
rejecting  that  denominator  and  retaining  only  the  numerator. 

It  should  always  be  an  object  with  the  scholar  to  preserve  the 
terms  of  a  fraction  as  small  as  is  possible  and  express  the  true 
value.  This  was  not  regarded  in  the  above  examples.  A  little 
experience  will  show  that  to  increase  or  diminish  the  value  of 
a  fraction,  it  is  only  necessary  to  make  the  numerator  larger  or 
smaller  compared  with  the  denominator.  Suppose  it  be  required 
to  multiply  the  fraction  1  by  2.  By  the  above  rule  the  product 
would  be  f,  which  is  equal  in  value  to  ^,  and  this  is  at  once 
obtained  by  dividing  the  denominator  by  2  instead  of  multiplying 

the  numerator  as  above,  thus,   -     _-   ;   therefore, 


A  fraction  is  multiplied  by  a  whole   number,  by  dividing   the 
denominator  by  that  number. 

The  following  examples  will  illustrate  this  principle  : 

1.  Multiply  |  by  2.     Ans.  f  . 

2.  Multiply  f  by  3.     Ans.  f  or  1. 

3.  Multiply  f  by  4.     Ans.  f  . 

4.  Multiply  T9^  by  5.     Ans.  f  . 

5.  Multiply  j3-  by  8.     Ans.  f  . 

6.  Multiply  -£T  by  7.     Ans. 

7.  Multiply 

8.  Multiply 

9.  Multiply 
10.  Multiply 


120  VULGAR  FRACTIONS. 

The  value  of  a  fraction  may  therefore  be  multiplied  by  a  whole 
number,  either  by  multiplying  the  numerator  or  dividing  the 
denominator  by  that  number. 

Note. — The  denominator  should  always  be  divided,  when- 
ever it  can  be  done  without  a  remainder. 

5.  A  fraction  is  divided  by  a  whole  number,  by  dividing  the 
numerator  by  that  number. 

This  needs  no  explanation.  If  we  divide  a  number  by  2, 
we  take  a  half,  and  if  by  3,  a  third  of  that  number  ;  that  is,  the 
divisor  always  shows  what  part  of  the  dividend  is  taken  ;  there- 
fore ,  if  H-  2  =  T6^,  and  if  ~  3  =  ^. 

The  following  examples  will  illustrate  the  operation  of  the 
above  principle : 

Ex.  1.  Divide  f  by  3.    Ans.  \. 

2.  Divide  f  by  2.    Ans.  f . 

3.  Divide  T8j  by  8.    Ans.  ^. 

4.  Divide  ^  by  6.     Ans.  ^. 

5.  Divide  if-  by  4. 

6.  Divide  |f  by  5. 

7.  Divide  -|  by  7. 

In  this  last  example,  the  scholar  will  find  a  difficulty.  He 
cannot  divide  the  numerator  in  any  way,  except  to  place  it  over 
the  7  in  the  form  of  a  fraction,  as  will  hereafter  be  explained  ; 
this  would  make  one  fraction  the  numerator  of  another 
fraction.  When,  therefore,  the  divisor  will  not  divide  the  nu- 
merator without  a  remainder,  a  more  convenient  mode  of  ope- 
rating is  desirable.  It  will  be  remembered,  that  division  is  the 
reverse  of  multiplication  ;  and  since  we  can  multiply  fractions 
by  dividing  the  denominator,  we  will  try  the  effect  of  dividing 
fractions  by  multiplying  the  denominator.  Let  it  be  required  to 
divide  T3^  by  3.  By  dividing  as  above,  we  obtain  T\>  as  the 
quotient,  viz.  ^  —  3=^.  By  the  mode  we  propose  to  try,  we 
obtain  ^.  It  therefore  remains  to  show  that  jV— A-  **  an7 
object  be  first  divided  into  12  equal  parts,  and  then  each  of 
these  12  parts  be  divided  into  3  equal  parts,  it  is  plain  that  the 
whole  would  be  divided  into  36  equal  parts,  and  that  each  twelfth 
part  would  make  3  thirty-sixth  parts ;  therefore,  y1^— ^ ;  hence, 
a  fraction  is  divided  by  a  whole  number  by  multiplying  its  de- 
nominator by  that  number. 


VULGAR  FRACTIONS.  121 

This  principle  may  be  applied  to  the  following  sums  : 
Ex.  1.  Divide  f  by  6.    Ans. 

2.  Divide  |  by  2.     Ans. 

3.  Divide  £  by  3.     Ans. 

4.  Divide  ^  by  5.     Ans. 

5.  Divide 

6.  Divide 

7.  Divide 

8.  Divide  }\ 

By  uniting  the  two  preceding  principles,  we  have  the  fol- 
lowing more  comprehensive  principle,  viz. : 

The  value  of  a  fraction  is  divided  by  a  whole  number,  by  di- 
viding the  numerator,  or  by  multiplying  the  denominator  by  that 
number. 

Note. — The  numerator  should  always  be  divided,  when  it 
can  be  done  without  a  remainder.  In  all  other  cases  the  de- 
nominator should  be  multiplied. 

From  the  preceding  remarks  and  illustrations  we  learn,  that 
whatever  operation  is  performed  on  the  numerator  of  a  fraction, 
the  SAME  OPERATION  is  PERFORMED  on  the  VALUE  of  the  frac- 
tion ;  but  that  the  effect  produced  on  the  VALUE  of  any  fraction,  is 
the  REVERSE  of  the  OPERATION  PERFORMED  ON  ITS  DENOMINATOR. 

6.  A  fraction  is  multiplied  by  a  fraction,  by  multiplying  the 
numerators  together  for  a  new  numerator,  and  the  denominators 
for  a  new  denominator. 

For  example,  let  it  be  required  to  multiply  £  by  \ .  Agree- 
ably to  the  principles  already  explained,  if  I  multiply  the  de- 
nominator of  the  fraction  \,  by  4,  the  other  denominator,  I  shall 
obtain  \  of  that  quantity,  viz.  ^  ;  and  if  I  multiply  this  quan- 
tity, viz.  J^,  by  3,  the  other  numerator,  I  shall  make  this  value 
three  times  as  large,  that  is,  it  will  become  -py ;  therefore,  -^ 
is  |  of  I,  or  £xf  =  TV 

In  accordance  with  the  above,  the  scholar  may  multiply  the 
following  fractions  : 

1 .  Multiply  £  by  J.    Ans.  -f^. 

2.  Multiply  %  by  f .     Ans.  if. 

3.  Multiply  f  by  J.    Ans.  ||. 

4.  Multiply  £1  by  f .    Ans.  ^. 

11 


122  VULGAR  FRACTIONS. 

5.  Multiply 

6.  Multiply 

7.  Multiply 

8.  Multiply 

9.  Multiply 
10.  Multiply 

7.  A  fraction  is  divided  by  another  fraction,  by  inverting  the 
divisor,  and  multiplying  them  together  as  before. 

A  unit  is  contained  in  the  fraction  },  three  fourths  of  once  ; 
^  is  consequently  contained  in  the  same  fraction  twice  as  often, 
viz.  f  of  a  time  ;  and  £,  three  times  as  often,  viz.  f  of  a  time, 
which  fractions  are  obviously  obtained  by  multiplying  f  by  -$ 
and  £  inverted.  Again,  suppose  it  be  required  to  find  how  many 
times  |  is  contained  in  $.  As  before,  a  unit  or  1  is  contained 
in  $,  seven  eighths  of  a  time  ;  £  would  be  contained  in  it  four 
times  as  often,  viz.  2¥8  of  a  time,  and  f  would  be  contained  in 
the  same  only  one  third  as  often  as  £,  viz.  -||  of  a  timez=l^4T,  or 
li.  This  result  is  obtained  by  inverting  the  divisor  |,  and 
multiplying  it  into  the  dividend  -g-  ;  thus,  ^xf=ff. 

The  following  examples  may  now  be  performed : 

Ex.  1.  Divide  £  by  f.    Ans.  f. 

2.  Divide  f  by  £.    Ans.  |-§. 

3.  Divide  |  by  ^.    Ans.  |f 

4.  Divide  f  by 

5.  Divide  TV  by  f . 

6.  Divide  ^  by 

7.  Divide  |  by  f 

8.  Divide  f  by  T9r.  • 

9.  Divide  4-  by  J. 
10.  Divide  {f  by  i. 

8.  If  the  numerator  and  denominator  of  any  fraction  be 
both  multiplied  or  both  divided  by  the  same  number,  the  value  of 
the  fraction  will  not  be  altered. 

Of  this  principle  no  explanation  is  necessary.  The  value 
of  the  fraction  being  the  quotient  arising  from  dividing  the 
numerator  by  the  denominator,  it  is  obvious  that  if  both  the 
terms  be  doubled,  or  repeated  any  number  of  times,  the  value  of 
the  quotient  will  not  be  affected. 


REDUCTION  OF  FRACTIONS.  123 


REDUCTION    OF    FRACTIONS. 

CASE  1st. — To  REDUCE  FRACTIONS  TO  THEIR  LOWEST  TERMS; 

OR,  TO  FIND  THE  LOWEST    TERMS  BY  WHICH    THE    VALUE  OF 
A  GIVEN  FRACTION  CAN  BE  EXPRESSED. 

RULE. — Divide  both  numerator  and  denominator  by  any  num- 
ber that  will  divide  them  both  WITHOUT  REMAINDER  ;  then  di- 
vide the  quotients  obtained,  in  the  same  manner,  and  so  continue 
to  do  till  there  is  no  number  greater  than  I,  that  will  divide  them. 
The  last  quotient  will  be  the  numerator  and  denominator  required. 

Ex.  1.  Reduce  |-f  to  its  lowest  terms.  Operation,  ||-f-8=|, 
and  J--K2  —  f ,  its  lowest  term. 

2.  Reduce  j-f^to  its  lowest  terms.     Ans.  |^. 

3.  Reduce  Jyff  to  its  lowest  terms.     Ans.  $. 

4.  Reduce  ^^  to  its  lowest  terms.     Ans.  £ . 

5.  Reduce  |§  to  its  lowest  terms.     Ans.  I. 

6.  Reduce  -fffo  to  ^ts  l°west  terms.     Ans.  Jj. 

7.  Reduce  |-§  to  its  lowest  terms.     Ans.  ^-. 

8.  Reduce  Q7\6ssj  to  its  lowest  terms.     Ans.  J. 

9.  Reduce  ^ffrfz  to  *ts  l°west  terms.     Ans.  %. 

10.  Reduce  6677886GQ-  to  its  lowest  terms.     Ans. 

11.  Reduce  y^ny  to  its  lowest  terms.     Ans. 

12.  Reduce  l-££  to  its  lowest  terms.     Ans.  l-f 

CASE  2d. — To  REDUCE  A  WHOLE  NUMBER  OR  A  MIXED   QUAN- 
TITY TO  AN  IMPROPER  FRACTION. 

RULE. — If  the  given  quantity  be  a  whole  number,  multiply  it 
by  the  proposed  denominator ;  the  product  will  be  the  numerator : 
but  if  it  be  a  mixed  quantity,  multiply  the  whole  number  by  the 
denominator  of  the  fraction,  and  to  the  product  add  the  given 
numerator ;  then  under  the  number  thu&  produced,  write  the 
denominator. 

Ex.  1.  Reduce  21  to  a  fraction  whose  denominator  is  9. 
Operation,  21  x9=;189,  the  numerator;  the  fraction  therefore 
is  i|^. 


124  REDUCTION  OF  FRACTIONS. 

2.  Reduce  8£  to  an  improper  fraction.    Operation,  8x3=24, 
and  24+1=25,  the  numerator  ;  therefore,  2^  is  the  answer. 

3.  Reduce  16|  to  an  improper  fraction.     Arts. 

4.  Reduce  17-^f  to  an  improper  fraction.     Ans. 

5.  Reduce  47^  to  an  improper  fraction.     Ans. 

6.  Reduce  135^  to  an  improper  fraction.     Ans.  --§--. 

7.  Reduce  1^|  to  an  improper  fraction.     Ans.  ^. 

8.  Reduce  1728-|y  to  an  improper  fraction.     Ans.  4-6-£j-  • 

9.  Reduce  9|to  an  improper  fraction.     Ans.  l-f. 

10,  Reduce   12|-  to  an  improper  fraction.     Ans.  y  . 

11.  Reduce  8  to  a  fraction  whose  denominator  shall  be  9. 


12.  Reduce  16  to  a  fraction  whose  denominator  shall  be  12. 
Ans.  ¥£. 

CASE  3d.  —  To   REDUCE  AN  IMPROPER  FRACTION  TO  A   WHOLE 

OR  MIXED  NUMBER. 

RULE.  —  Divide  the  numerator  by  the  denominator;  the  quo- 
tient will  be  the  whole  number.  If  there  be  any  remainder,  place 
it  over  the  denominator  at  the  right  of  the  whole  number. 

Note.  —  The  true  quotient  includes  both  the  whole  number 
and  fraction.  In  all  cases  of  division,  therefore,  the  remainder 
(if  any)  constitutes  the  numerator  of  a  fraction  of  which  the 
divisor  is  the  denominator. 

Ex.  1.  Reduce  Yy1  to  a  mixed  number. 

4  OPERATION. 

17)141(8 
1  3  6 

5  rem.  ;  therefore,  8T5y  is  the  answer. 

2.  Reduce  *•£•£  to  a  mixed  quantity.     Ans. 

3.  Reduce  7^9  to  a  mixed  quantity.    Ans.  13-J. 

4.  Reduce  *-f  to  a  mixed  quantity  or  whole  number.     Ans.  7. 

5.  Reduce  4-ff  to  a  mixed  quantity.     Ans.  5f  1. 

6.  Reduce  ^^4  to  a  mixed  number.     Ans. 

7.  Reduce  ^  to  a  mixed  number.     Ans. 

8.  Reduce  -f4-  to  a  mixed  number.     Ans.  56-37T. 

9.  Reduce1!^  to  a  whole  number.     Ans.  12. 
10.  Reduce  ff  to  a  mixed  number.     Ans.  6^. 

1  1  .  Reduce  l-7-g-8  to  a  whole  or  mixed  number.     Ans.  288, 
12.  Reduce  5-°27T8-9  to  its  proper  number.     Ans.  2704^-. 


REDUCTION  OF  FRACTIONS.  125 


CASE  4th. — To  REDUCE  COMPOUND  FRACTIONS  TO  SIMPLE  ONES. 

RULE  1st. — Multiply  all  the  numerators  together  for  a  new 
numerator,  and  all  the  denominators  for  a  new  denominator, 
and  reduce  the  new  fraction  to  its  lowest  terms,  by  Case  1st. 

Ex.1.  Reduce  f  of  |  of  f  to  a  simple  fraction.  Performed, 
2  X  3  X  5  =  30,  the  new  numerator ;  and  3  X  4  x  6  =  72,  the  new 
denominator  ;  therefore,  |-§  is  the  fraction  required,  but  suscepti- 
ble of  being  expressed  in  lower  terms  ;  therefore,  j^-^^=Y5i 
Ans. 

Compound  fractions  may  be  reduced  to  simple  ones,  however, 
much  more  expeditiously,  by  canceling.  The  labor  of  reducing 
to  lower  terms  is  thereby  avoided. 

RULE. — Draw  a  horizontal  line  and  place  all  the  numerators 
above  the  line  and  all  the  denominators  below  it.  Cancel  the 
numbers  as  far  as  practicable,  as  taught  in  the  Rule  for  Can- 
celing ;  then  make  the  product  of  the  numbers  remaining  above 
the  line  the  new  numerator,  and  the  product  of  those  remaining 
below,  the  new  denominator. 

Note  1st. — If  there  be  nothing  remaining  above  the  line  after 
canceling,  1  will  alw^s  be  the  numerator  of  the  new  fraction. 
The  same  is  true  of  the  denominators. 

Ex.  2.  Reduce  £  of  f  of  |  to  a  simple  fraction. 

Statement,  ^  ^  &.     Canceled,  g'  4  &.     Ans.   ^. 

Example  1st  stated  and  solved  by  canceling : 
2.  3.  5       ~         ,  j  2.  $.   5 

m-    Canceled>  riTe-    Ans'  A- 

2 

3.  Reduce  f  of  ^  of  -J J  to  a  simple  fraction. 

Statement,  %±£.     Canceled,  |i|*      Ant.  i. 

4 

Note  2d. — Whenever  the  product  of  any  two  numbers  on 
one  side  of  the  line  will  cancel  any  number  on  the  opposite 
side,  they  may  be  so  canceled  ;  as  in  the  last  example,  7  and  2 
below  the  line  cancel  14  above  it. 

4.  Reduce  |  of  ^J  of  |  of  |  to  a  simple  fraction. 


126  REDUCTION  OF  FRACTIONS, 

Statement,  |-^.     Canceled,  |^|£| 

3  2 

and   7  x2  m  14,  numerator  ;  and  13  x  3  =  39,  denominator  ; 
therefore,  the  new  fraction  is  J£. 

5.  Reduce  f  of  lf  of  T^  of  T\  of  T7^  to  a  simple  fraction. 


6.  Reduce  i  of  |  of  f  of  f  of  f  of  f  of  |  of  £  of  T%  to  a  simple 
fraction.     .Arcs.  Jg.. 

7.  Reduce  if  of  f  of  T9^  of  f  of  f  of  f  to  a  simple  fraction. 
An*.  f-ff  . 


.  —  If  any  term  of  a  compound  fraction  be  a  mixed 
number,  it  must  be  reduced  to  an  improper  fraction  before 
stating. 

8.  Reduce  ^  of  ^  of  4f  of  f  of  f  to  a  simple  fraction. 
4f  —  y,  therefore,  statement,    '    '      '    '-^  ;    which   canceled 

will  give  the  answer,  J. 

9.  Reduce  f  of  7£  of  J  of  2|-  to  a  simple  fraction.     Ans. 

21  6  _  f\  6 
3T  —  °3^J' 

10.  Reduce  £  of  |  of  ^  of  J  of  |-  to  a  simple  fraction.     Ans. 

.  1  6 

11.  Reduce  T9^  of  ^  of  ^  to  a  simpl^fraction.     JL»^.  -Jf. 

12.  Reduce  -f^  of  J  of  J  of  j1^  to  a  simple  fraction.     .Aws.  ^. 

13.  Reduce  ^  of  i|  of  |^  to  a  simple  fraction.      Ans. 

14.  Reduce  f  of  J  of  ^  of  i  to  a  simple  fraction.     ^In. 

15.  Reduce  f  of  8|  of  5|-  to  a  simple  fraction.     Ans.  ^ 
or  701. 

CASE  5th.  —  To  CHANGE  FRACTIONS  FROM  ONE  DENOMINATION  TO 

ANOTHER,  WITHOUT  ALTERING  THE  VALUE. 

1st.  To  reduce  fractions  of  low  denominations  to  those  of 
higher  value. 

RULE.  —  Divide  the  fraction,  or  what  is  the  same  thing,  mul- 
tiply the  denominator  by  such  numbers  as  are  required  to  reduce 
the  given  quantity  from  the  GIVEN  to  the  REQUIRED  DENOMI- 
NATION. 

Ex.  1  .  Reduce  ^  of  a  penny  to  the  fraction  of  a  pound.  The 
numbers  required  to  reduce  pence  to  pounds,  are  ]  2  and  20  ; 
therefore,  f  of  a  penny  is  to  be  divided  by  these  numbers  ;  and 
since  this  can  be  effected  in  the  present  case  only  by  multiplying 


REDUCTION  OF  FRACTIONS.  127 

the  denominator,  the  operation  will  be,  ^     J2     2Q  =  j^,  and 

this,  by  Case    1st,  is  reduced  to   yj^,   which  is  the  fraction 
required.     Hence  f  of  a  penny  equals  ^|-g  of  a  pound. 

The  canceling  principle  may  however  be  successfully  ap- 
plied in  the  solution  of  sums  of  this  character. 

RULE  FOR  CANCELING.  —  Place  the  numerator  of  the  given 
fraction  above  a  horizontal  line,  and  its  denominator  below  it  ; 
then  place  also  BELOW  the  line,  such  numbers  as  are  necessary  to 
reduce  the  denomination  given  to  that  required.  Cancel,  tyc. 
as  before. 

We  will  solve  the  above  example  by  this  rule  also.    Statement, 

2       The  scholar  should  compare  the  statement  with  the 

rule,  to  see  that  he  understands  its  application.     The  above 

statement  canceled,      10       ;  and  6  x  12  x  4=288,  the  denomi- 
o.  i<«.  iJU 
^4 

nator  as  before,  and  nothing  remains  as  a  numerator  ;  therefore, 

as  before,  ^-i-g-  of  apoundis  the  answer.  (See  Note  1st,  Case  4th.) 

2.  Reduce  f  of  a  farthing  to  the  fraction  of  a  shilling.     By 

3  3 

the   common   rule,  jx4xl2_  ^»  which,   by  Case  1st,  equals 

3 
,  Ans.     By  the  rule  for  canceling,  ^        ^.     The  same  can- 

4.  4. 


celed'  =&*   Ans' 


4 

3.  Reduce  £  of  a  penny  to  the  fraction  of  a  pound.     Statement, 

5~l2~20-     Canceled'  5  12  20  ;  and  5  X  3  X  20  =  300,  therefore, 

3 
Ans.  g^j. 

4.  Reduce  £  of  a  gallon  to  the  fraction  of  a  hogshead.    Ans. 

Statement,  - — 7^.     The  63  below  the  line  reduces  the  gal- 
lons to  hogsheads. 

5.  Reduce  f  of  an  ounce   Troy  to  the  fraction  of  a  pound. 
Ans.  3^. 

6.  Reduce  \^  of  a  minute  to  the  fraction  of  a  day.     Ans. 

T~5  3"(K 

7.  Reduce  T8T  of  a  pound   Avoirdupois  to  the  fraction  of  a  ' 
cwt.     Ans. 


128  REDUCTION  OF  FRACTIONS. 

8.  Reduce  f  of  a  nail  to  the  fraction  of  an  ell  English.  Ans. 

3 

160* 

9.  Reduce  T5^  of  a  penny  to  the  fraction  of  a  pound.     Ans. 


10.  Reduce  i-J  of  an  hour  to  the   fraction  of  a  year.     Ans. 

9T9"0' 

Secondly.  To  reduce  fractions  of  high  denominations  to 
equivalent  fractions  of  lower  denominations. 

RULE. — Multiply  the  numerator  by  such  numbers  as  are  re- 
quired to  reduce  the  given  quantity  from  the  given  to  the  re- 
quired denomination,  and  then  by  Case  1st  reduce  the  result  to 
its  lowest  terms. 

Ex.  1 .  Reduce  ^  of  a  shilling  to  the  fraction  of  a  farthing. 

To  reduce  shillings  to  farthings,  we  must  multiply  by  12 
and  4;  therefore,  ^g-xl2x4=: |-|;  and  by  Case  1st,  ff^i, 
Ans. 

RULE  FOR  CANCELING. — Place  the  numerator  of  the  given 
fraction  above  a  horizontal  line,  and  the  denominator  below,  as 
before  ;  then  place  above  the  line  such  numbers  as  are  necessary 
to  reduce  the  denomination  given  to  that  required.  Cancel,  <SfC. 
as  before. 

1     12.  4 

The  above   sum  solved  by  this  rule.     Statement,  -r — 

1    12    4 
The  same  canceled,     — • — ;  therefore,  ^  of  a  farthing  is 

&        2 
the  answer. 

The  scholar  will  carefully  observe  the  difference  between 
the  statement  here,  and  the  one  given  for  reducing  low  denom- 
inations to  high. 

2.  Reduce  -^4^  of  a  pound  to  the  fraction  of  a  penny.  Ans.  f . 

1.  20.  12 
Statement,    — • . 

3.  Reduce  -jig  of  a  pound  Troy  to  the  fraction  of  a  pwt. 

•I       -jo      Of) 

Ans.  2^°,  or  2}  pwt.      Statement,  -^ — 

4.  Reduce  y1^-  of  a  pound  Troy  to  the  fraction  of  an  ounce. 
Ans.  f . 


REDUCTION  OF  FRACTIONS.  129 

5.  Reduce   ^  of  a   penny  to   the  fraction  of  a  farthing. 
Ans.  |. 

6.  Reduce  Q6Q1000  of  a  mile  to  the  fraction  of  a  barley  corn. 

Ans-  ^jV 

7.  Reduce  yij  of  a  cwt.  to  the  fraction  of  a  pound  Avoirdu- 
pois.    Ans.  -j^-. 

8.  Reduce  yf^  of  an  ell  English  to  the  fraction  of  a  nail. 
Ans.  f . 

9.  Reduce  94^0  °f  a  year  to  t^ie  fraction  of  an  hour.     Ans. 
if- 

CASE  6th. — To  REDUCE  FRACTIONS  OF  A  HIGHER  DENOMINA- 
TION TO  THEIR  VALUE  IN  WHOLE  NUMBERS  OF  A  LOWER 
DENOMINATION. 

RULE. — Reduce  the  fraction  to  its  next  lower  denomination  by 
multiplying  the  numerator  by  the  requisite  number,  and  divide 
the  product  by  the  denominator ;  the  quotient  thus  obtained  will 
be  a  whole  number  of  the  lower  denomination,  and  the  remainder, 
if  any,  may  be  reduced  and  divided  as  before.  This  process 
may  be  continued  till  nothing  remains,  or  till  it  is  reduced  to  the 
lowest  denomination. 

Ex.  1.  Reduce  f  of  a  pound  sterling  to  its  value  in  shillings, 
and  pence. 

OPERATION. 
2= number. 
2  0 

Div.  by  denoin.     3)4  Omshillings. 

13s.  and  1  s.  remains,  which  equals  12  d., 
and  12  d.-^-3=4  d.  Therefore,  13  s.  4  d.  is  the  required  num- 
ber. 

It  is  evident  that  f  of  a  pound  sterling  is  20  times  as  many 
thirds  of  a  shilling;  viz.  4^°  =  13£  shillings  ;  and  ^  of  a  shil- 
ling is  12  thirds  of  a  penny,  that  is,  L2=:4d.  Hence,  f  of  a 
pound  is  13  s.  4  d. 

2.  Reduce  -^  of  a  pound  sterling  to  its  value  in  lower  denom- 
inations. 

Solution  :  Jg-  of  a  pound =|$  of  a  shilling,  and  £§  of  a  shil- 
2^°vof  a  penny=4d.  Ans. 

3.  Reduce  f  of  a  pound  Troy  to  its  integral  value.  Ans.  9  oz. 


130  REDUCTION  OF  FRACTIONS. 

4.  Reduce  ^g-  of  a  day  to  its  integral  value.     Ans.  1  h.  20  m. 

5.  Reduce  -£$  of  an  hour  to  its  value   in   whole  numbers. 
Ans.  54  m. 

6.  Reduce  ^  of  a  hogshead  to  its  value  in  whole  numbers. 
Ans.  2  qt.  1  pt.  1  gi. 

7.  Reduce  f  of  a  quart  to  its  integral  value.     Ans.  1  pt.  1  gi. 

8.  Reduce   f  of  an  ell  English  to   a  whole  number.     Ans. 
3qr. 

9.  Reduce  f  of  a  yard  to  a  whole  number.     Ans.  3|qr. 
10.  Reduce  f  of  an  ell  French  to  a  whole  number.     Ans. 

4qr. 

CASE  7th.— To  REDUCE  THE  LOWER  DENOMINATIONS  OF  A 

COMPOUND  NUMBER  TO   FRACTIONS  OF  A  HIGHER   DE- 
NOMINATION. 

RULE.-— Reduce  the  given  quantity  to  the  lowest  denomination 
in  that  quantity,  for  a  numerator  of  the  fraction  ;  and  then  re- 
duce a  unit  of  the  higher  denomination  to  the  same  denomination 
with  the  numerator  for  the  denominator  of  the  fraction.  This 
fraction  reduced  to  its  lowest  terms  by  Case  1st,  will  be  the  one 
required. 

Ex.  1 .  Reduce  2  qr.  2  na.  to  the  fraction  of  a  yard. 

Operation,  2qr.  2  na.r=  10  nails,  the  numerator  ;  and  lyd. 
r=4qr.,  and  4qr.  =  16  nails,  the  denominator;  therefore,  yf  is 
the  fraction,  which  equals  f ,  Ans. 

The  operation  may  be  much  abbreviated  by  canceling  f  for 
which  the  following  will  be  found  a  convenient  rule. 

RULE  FOR  CANCELING. — Reduce  the  given  quantity  to  the 
lowest  denomination  mentioned,  (if  it  consist  of  different  denom- 
inations,) and  place  it  above  a  horizontal  line  ;  and  beneath  the 
same  line  place  the  numbers  required  to  reduce  this  denomination 
to  the  required  denomination.  Cancel,  multiply,  fyc.,  and  the 
terms  of  the  required  fraction  will  be  obtained. 

Ex.  2.  Reduce  3s.  4  d.  to  the  fraction  of  a  pound. 
Operation,    3s.  4  d.  =  40  d.,  and  pence  are  reduced  to  pounds 
by  dividing  by  12  and  20.     Therefore, 

Statement,  -         .     Canceled, 


20  '»  13. 


REDUCTION  OF  FRACTIONS.  131 

Nothing  remains  above  the  line,  the  numerator  of  the  required 
fraction  therefore  is  1  ;  and  6  remains  below  the  line,  and 
consequently  is  the  denominator  ;  therefore,  1  is  the  fraction 
required,  and  3s.  4  d.  is  ^  of  a  pound. 

3.  Reduce  2  roods   and  30  rods  to  the  fraction  of  an  acre. 

2  roods  and  3  rods  =  110  rods;  and  rods  are  reduced  to 
acres,  by  dividing  by  40  and  by  4  ;  the  statement  therefore  is, 
110  ,  ,  .  im.  „  A 

Same  cance*ed  1S>  ~~~> 


40~T' 

4.  Reduce  12  ounces   to  the  fraction  of  a  pound  Avoirdu- 
pois.    Ans.  f  . 

5.  Reduce  26  gal.  2  qt.  to  the  fraction  of  a  hogshead.    Ans. 

T2\' 

6.  Reduce  3  fur.  20  rods  to  the  fraction  of  a  mile.    Ans.  -f^. 

7.  Reduce  3  qr.  2  na.  to  the  fraction  of  an  ell  English.    Ans. 

TV 

8.  Reduce  2  dr.  2  sc.  to  the  fraction  of  an  ounce.     Ans.  $. 

9.  Reduce  2  qr.  24  Ib.  to  the  fraction  of  a  cwt.     Ans.  |-. 

10.  Reduce  6  oz.  lOpwt.  to   the   fraction  of  a  pound  Troy. 
Ans.  Jf. 

11.  Reduce  6qt.  to  the  fraction  of  a  bushel.     Ans.  ^. 

12.  Reduce  12h.  30m.  to  the  fraction  of  a  day.     Ans.  J|. 

13.  Reduce  4  d.  12h.  to  the  fraction  of  a  week.     Ans^  T9T. 

14.  Reduce  15  deg.  30  m.  to  the  fraction  of  a  Sign.     Ans. 

W 

CASE   8th.  —  To  REDUCE  FRACTIONS  HAVING  DIFFERENT   DE- 

NOMINATORS TO  EQUIVALENT  FRACTIONS  HAVING  A 
COMMON  DENOMINATOR. 

RULE.  —  Multiply  all  the  denominators  together  for  a  new 
denominator,  and  each  numerator  into  all  the  denominators  ex- 
cept its  own,  for  a  new  numerator  to  each  fraction.  The  several 
numerators  placed  over  the  common  denominator  will  give  the  re- 
quired fractions. 

Note.  —  The  fractions  should  be  reduced  to  their  lowest 
terms  before  multiplying.  If  the  scholar  looks  carefully  into 
the  nature  of  this  rule,  he  will  see  that  it  is  only  multiplying 
numerators  and  denominators  by  the  same  numbers  ;  and  he 
has  already  learned,  that  this  does  not  affect  the  value  of  the 
fraction. 

Ex.  1.  Reduce  -|  ,  £,  and  -^-,  to  a  common  denominator. 


132 


ADDITION  OF  FRACTIONS. 


PERFORMED. 

7x5x1  1  =  385,  the  common  denominator. 

6x5x11  =  330,  the  numerator  for  f,  which  therefore  equals 


4x7x1  1  =  308,  the  numerator  for  f  ,  therefore,  £= 
8  X  5  X  7  =  280,  the  numerator  for  ^,  therefore,  j^^ff  °  . 
The  required  fractions,  therefore,  are  f| 


2.  Reduce 


.  and 


,  to   a  common   denominator.      Ans. 


3.  Reduce  -|  ,  f  ,  J,  and  •§•  ,  to  a  common  denominator.     Ans. 

168       72         84        63 
2~52>    252>  ^5  2>  252* 

4.  Reduce  f  ,  Y  ,  T^,  and  -f^,  to  a  common  denominator.  Ans. 

230J:      1JL5_6_P     _2_4JL    UilO. 
2880'  '2880  '    2880'    2880' 

5.  Reduce  |,  f  ,  |-  ,  and  ^,  to  a  common  denominator.    Ans. 

1080     1440       486 
'  T6~2"0'  T6^¥'    1620' 

6.  Reduce  f,  J,  |,  and  T8T  to  a  common  denominator.     Ans. 

891    495   1155   1080 
T4¥5'  TT8  5"'  TT8T'  T4¥5* 

At  the  commencement  of  this  rule,  the  scholar  was  instruct- 
ed relative  to  the  peculiar  form  and  nature  of  fractions,  and 
made  acquainted  with  certain  principles  of  universal  applica- 
tion. In  the  course  of  the  preceding  eight  cases,  he  has  been 
shown  the  various  changes  of  which  fractions  are  susceptible, 
while  their  value  remains  unaffected.  His  attention  will  now 
be  directed  to  those  operations  by  which  their  value  is  affected. 


ADDITION    OF   FRACTIONS. 

If  the  scholar  will  turn  back  to  Simple  Addition,  he  will 
there  find  it  stated,  that  numbers,  or  quantities  of  the  same 
kind  only,  can  be  reduced  to  a  single  number  or  quantity  by 
adding.  The  same  is  true  of  fractions.  It  is  obvious  that  ^ 
of  a  shilling,  and  |  of  a  penny,  make  neither  f  of  a  shilling 
nor  |  of  a  penny.  But  £  of  a  shilling  makes  ^  of  a  penny ; 
and  to  this,  we  can  add  £  of  a  penny,  and  the  amount  will  be 
L3  of  a  penny.  Therefore,  before  we  can  add  fractions,  they 
must  be  reduced  to  the  same  denomination.  (See  Case  5th.) 


ADDITION    OF    FRACTIONS.  133 

It  is  equally  impossible  to  add  fractions  whose  denominators 
are  unlike.  £  of  a  shilling  added  to  £  of  a  shilling,  makes 
neither  J  of  a  shilling  nor  J  of  a  shilling.  But  £  of  a  shilling 
=  f  of  a  shilling  ;  and  f  4~i  =f  of  a  shilling.  Fractions  must 
therefore  be  reduced  to  a  common  denominator,  before  they  can 
be  united.  (See  Case  8th.)  Hence  we  have  the  following  rule. 

RULE.  —  Reduce  all  the  fractions  to  the  same  denomination, 
and  also  to  a  common  denominator  ;  then  ado"  their  numerators 
and  place  their  sum  over  the  common  denominator.  If  the  frac- 
tions produced  be  improper,  reduce  them  to  a  whole  number  or 
mixed  quantity. 

Note.  —  If  any  of  the  fractions  are  compound,  they  must  be 
reduced  to  simple  ones,  before  they  can  be  reduced  to  a  com- 
mon denominator.  (See  Case  4th.) 

Ex.  1.  What  is  the  sum  of  ^  and  f-?  These  fractions,  reduced 
to  a  common  denominator  by  Case  8th,  become  -^  and  ^-J,  and 


2.  What  is  the  sum  of  f  of  J,  and  f  of  j-  ?     By  Case  4th,  £ 
of  J=TV  and  |  of  £=£,  and   *  +4=^+^,  (see   Case 

8th,)  and  ^+^=11^1^. 

3.  What  is  the  sum  of  ^  and  -|  ?     Ans.  y  or  If. 

4.  What  is  the  amount  of  £  of  J  and  f  of  i|  ?     Ans.  fj. 

5.  What  is  the  amount  of  18  and  16,  and  £  of  %  1      Ans. 
34f. 

Note  2d.  —  When  whole  numbers  are  combined  in  the  same 
operation  with  fractions,  add  each  separately  and  unite  them, 
as  in  the  above  sum. 


6 .  What  is  the  amount  of  2 1 , 7,  £,  and  f  of  Y?\  A ns. 

7.  What  is  the  amount  of  f  of  a  penny  adde\to  ^  of  a  £>.  ? 

Explanation. — ^  of  a  £.  =  8^°  of  a  penny  ;  and  f  of  a  penny 
-f-8^0  of  a  penny  —  3j^  of  a  penny,  and  this  equals  2s.  3d. 
If  qr.  Ans. 

8.  What  is  the  amount  of  f  of  a  yard  and  £  of  a  nail  ?     Ans.  3 
qr.  04:  na. 

9.  What  is  the  sum  of  |-  of  a  pound  added  to  ^  of  a  shilling  ? 
Ans.  ^3s.  =  i7s.  2d. 

10.  What  is  the  sum  of  \  of  i  and  f  of  \\.    Ans.  |^  or  1-ff . 

11.  What  is  the  sum  of  £  of  f  of  f  off  of  f  off,  and  i  ? 
.Aws.  if 

12 


134 


SUBTRACTION  OF  FRACTIONS, 


12.  What  is  the  sum  of  J  of  a  ton  added  to  f  of  a  cwt.  1 
Ans.  llf^  cwt. 

13.  What  is  the  sum  of  f  of  a  day  added  to  $  of  an  hour  ? 
Ans.  19^  hours. 

14.  What  is  the  sum  of  -J  of  a  pound,  f  of  a  shilling,  and  £ 
of  a  penny  ?     Ans.  3  s.  2  d.  2  qr. 

15.  What  is  the  sum  of  -^  of  a  week,  f  of  a  day,  and  £  of  an 
hour  ?     Ans.  1  day,  22  hours,  15m. 


SUBTRACTION  OF    FRACTIONS. 


As  we  can  subtract  a  quantity  only  from  another  of  the 
same  kind,  it  is  obvious  that  the  same  preparations  are  neces- 
sary to  perform  operations  in  this  rule  as  in  the  preceding  ; 
therefore, 

RULE. — Prepare  the  fraction  as  in  addition,  then  subtract 
the  less  numerator  from  the  greater,  and  place  the  remainder 
over  the  common  denominator. 

It  will  be  obvious  that  the  difference  of  the  numerators  is 
the  difference  sought. 

Ex.  1.  From  |- take  ^.  Operation,  |  —  s^fi —  A— if> 
Ans. 

2.  From  f  take  £-.     f — •2T=Tft —  ~lA—i*/9'  Ans. 

3.  From  |-  take  |-.  Ans.  ^.     In  this  last  example  it  is  evident, 
that,  as  the  denominators  are  the  same,  the  operation  consists 
in  subtracting  the  numerators  only.     The  same  is  true  of  all 
similar  examples,  provided  only  that  the  fractions  are  of  the 
same  denomination. 

4.  From  11  take  -~.     Ans. 

5.  From  f|  take  if.     Ans. 

6.  From  if  take  ^.     Ans.  •, 

7.  From  f  f-  take  if.     Ans. 

8.  From  £  of  a  pound  take  £  of  a  shilling.     \  of  a  pound 

—  22°  of  a  shilling  :  and  2^°  =  6g°  of  a  shilling.     Therefore,    %° 

—  %=*•£  of  a  shilling,  and  5^  of  a  shilling =9  s.  2  d. 

9.  From  J  of  a  league  take  -£  of  a  mile.      1   league  =  3 


-  or    . 

or  f 
or  £ 


MULTIPLICATION  OF  FRACTIONS.  135 

miles  ;  therefore,  f  of  a  league =f  of  a  mile  ;  and  ^  of  a  mile 
=%  of  the  same  ;  hence,  J  — }=-j  or  1|  of  a  mile,  which  is 
the  distance  required. 

10.  From  £  of  a  shilling  take  f  of  a  penny.     Ans.  ^  or  5£ 
pence. 

11.  From  |  of  a  day  take  f  of  an  hour.     Ans.  1-f4-,  or  20-f- 
hours. 

12.  From  T%  of  a  pound  Avoirdupois  take  |  of  an  ounce. 
Ans.  Vu3>  or  13£$  ounces. 

13.  "From  J  take  1  of  f  of  f .     .4/i.y.  $.     $  of  f  of  f  nr^,  and 
S--i=aor|. 

14.  From  }i  of  an  ell  English  take  £  of  a  yard.     Ans.  f£ 
or  2TV  yards. 

15.~  From  9|  take  6|.     A/w.  3^. 

16.  From  19  yards  take  5|-  yards.     Ans.  13|. 

17.  From  7  Ells   English  take  4£  yards.     Ans.   Y=4i 
yards,  or  3|  Ells  English. 

1 8.  From  £  of  a  pound  sterling  take  f  of  a  penny.     Ans. 
3TV  of  a  penny,  or  2  s.  1 J4  d. 

"19.  From  f  of  a  rod  take  |  of  a  foot.  Ans.  y  or  10£  feet. 
20.  From  |-  of  an  ounce  take  |  of  a  pwt.  Ans.  Vj  pwt.  or 
pwt. 


MULTIPLICATION  OF   FRACTIONS. 

A  fraction  may  be  multiplied  by  a  whole  number,  either  by 
multiplying  the  numerator  or  dividing  the  denominator  by  that 
number.  This  has  been  fully  illustrated  in  Section  4th,  of 
Fractions. 

Ex.  1.  Multiply  fcby  16.     Ans.  V6,  or  5J. 

2.  Multiply  |  by  8.     Ans.  %°,  or  6|. 

3.  Multiply  f  by  9.     Ans.  5T4>  or  7f . 

4.  Multiply  f  by  3.     Ans.  ^4,  or  2f . 

5.  Multiply  SL*.  by  7.     _4/w.  2g°,  or  6f. 

6.  Multiply  |i  by  8.     ^TW.  2J,  or  7. 

7.  Multiply  ||  by  12.      Ans.  \3,  or  11. 

8.  Multiply  |f  by  3.     Ans.  2f ,  or  3f. 

A  fraction  is  multiplied  into  a  whole  number  equal  to  its  de- 
nominator, by  rejecting  that  denominator. 


136  MULTIPLICATION  OF  FRACTIONS. 

9.  Multiply  if  by  21.  By  dividing  if  by  21,  I  take  ^r 
part  of  15 ;  if,  then,  this  JT  part  be  repeated  21  times,  it  is 
evident  that  the  value  of  all  the  parts  will  equal  15. 

10.  Multiply  f|  by  82.     Ans.  72. 

11.  Multiply  I  by  9.     Ans.  7. 

12.  Multiply  4J  by  73.     Ans.  41. 

13.  Multiply  y  by  43.     Ans.  21. 

TO  MULTIPLY  FRACTIONS  BY  FRACTIONS. 

RULE. — Multiply  the  numerators  of  the  given  fractions  to- 
gether for  the  required  numerator,  and  the  denominators,  for 
the  required  denominator ;  then  by  Case  1st  reduce  the  terms 
as  far  as  practicable. 

Note. — Mixed  numbers  are  to  be  reduced  to  improper  frac- 
tions before  multiplying ;  or  we  may  first  multiply  the  integris 
and  then  the  fractions,  and  add  their  products. 

Ex.  1.  Multiply  |  by  f    |xf=ft,  and  ^-3=TV,  Ans. 

2.  Multiply  f  by  J .     f  x  J=H=&»  A™- 

3.  Multiply  7|by3£.     7$  =  \?,   and  3$  =  \j>,  and  ^5  X  V° 
—  I|Q_2_5j  or  25,  Ans. 

RULE  FOR  CANCELING. — Place  the  numerators  of  the  given 
fractions  above  a  horizontal  line,  and  their  denominators  be- 
low the  same.  Cancel,  multiply,  fyc.  as  before. 

One  important  advantage  of  the  above  rule  will  be  found  in 
the  fact,  that  it  always  gives  the  answer  in  its  lowest  possi- 
ble terms. 

4.  Multiply  |  by  f . 

1    o  13 

Statement,  ^— .     Canceled,  «— -=-nr,  Ans. 

o.  5  o,    5 

2 

5.  Multiply  |  of  |  of  f  by  |  of  f  of  J$. 

3.  1.  2.  1,  2.  11       ~         .    .   B.  1.  2.  1.  S.  11 
Statement,  ^^-3—--.     Canceled.  4  fi ^  &  g   ~  ; 

2 

therefore,  1 1  ^numerator,  and  2x6x9x12  =  1296,   denomi- 
nator, and  y^g-,  Ans. 

6.  Multiply  A  by  T\. 

9      3 
Statement, 


MULTIPLICATION  OF  FRACTIONS.  137 

7.  Multiply  i  by  2*.     Ans.  ff  ,  or  2T\. 

8.  Multiply  16£  by  12£.     Ans.  203^. 

9.  Multiply  13i  by  9|.     Ans.  124^. 

10.  Multiply  £  of  f  of  f  of  |  of  j-  of  f  of  £  of  f  by  £  of  T9F 
of      of      . 


In  solving  sums  by  canceling,  like  the  above,  the  necessity 
of  reducing  compound  fractions  to  simple  ones  is  avoided. 


11.  Multiply  $  of  19  by  f. 

1    19    3 
Statement,  g'    "*  6>     An*.  ^  ,  or 


If  any  whole  numbers  are  given  as  parts  of  the  dividend,  it 
is  only  necessary  to  write  them  above  the  line  as  in  the  last 
example  ;  or,  if  they  are  given  as  parts  of  the  divisor,  they 
only  require  to  be  placed  below  the  line  ;  the  operation  then 
proceeds  as  before. 

12.  Multiply  f  of  10  by  f  .     Ans.  3. 

13.  Multiply  144  by  T\.     Ans.  12. 

14.  Multiply  395  by  £  of  f  .     Ans.  *£§.,  or  52f  . 

15.  Multiply  f  of  3  times  T%  of  5  times  £  of  i  of  £,  by  4 
of^offof^.     Ans.  J. 

16.  What  will  2^  tons  of  hay  cost  at  16|  dollars  per  ton? 
Ans.  41  'i  dollars. 

17.  What  will  2^  barrels  of  sugar  cost  at  18£  dollars  per 
barrel  ?     Ans.  42-f^  dollars. 

18.  What  will  8f  pounds  of  tea  cost  at   1^   dollars  per 
pound?     Ans.  lO^f  dollars. 

19.  What  will  4f  cords  of  wood  cost  at  3f  dollars  per 
cord?     Ans.  1  7  1|  dollars. 

20.  What  will  9  yards  of  cloth  cost  at  %  dollar  per  yard  ? 
Ans.  7%  dollars. 

21.  What  will  34  gallons  of  wine  cost  at  If  dollars  per 
gallon  ?     Ans.  46^  dollars. 

22.  What  will  12£  barrels  of  sugar  cost  at  15^  dollars  per 
barrel?     Ans.  190f  dollars. 

23.  What  will   22£  pounds  of  lard  cost  at  J   dollar  per 
pound  ?     Ans.  2|£. 

12* 


138 


DIVISION  OF  FRACTIONS. 


DIVISION    OF    FRACTIONS. 

Division  of  Fractions  is  naturally  divided  into  the  three 
following  kinds  ;  viz.  the  division  of  a  fraction  by  a  whole 
number  ;  the  division  of  a  whole  number  by  a  fraction  ;  and 
the  division  of  a  fraction  by  a  fraction. 

Division  of  fractions  by  a  whole  number  was  fully  illustra- 
ted in  Sec,  5th  of  the  remarks  introductory  to  this  rule.  It  is, 
therefore,  necessary  here  merely  to  repeat,  that  a  fraction  is 
divided  by  a  whole  number,  either  by  dividing  its  numerator,  or 
multiplying  its  denominator  by  that  number. 

Ex.  1.  Divide  -Jf  by  9.     Ans.  £.  | 

2.  Divide  f  by  7.     Ans.  /j. 

3.  Divide  L3  by  11.     Ans.ffi. 

4.  Divide  Hb73-     Ans.  ^y. 

5.  Divide  Si  by  9.     Ans.  &,  or  &. 

6.  Divide  jL-  by  8.     Ans.  ^J^-. 

7.  Divide  fi-  by  6.     Ans.  %±. 
S.  Divide  y  by  5.     Ans.  £f 
9.  Divide  ^  by  12.     Ans. 

10.  Divide  £2  by  7.     Ans.  - 

11.  Divide  y>  by  40.     Ans. 

12.  Divide  J-  by  72.     Ans. 


It  is  obvious  that  the  quotient  arising  from  dividing  a  whole 
number  by  a  fraction,  must  be  as  much  larger  than  the  number 
itself,  as  a  unit  or  1  is  greater  than  the  fraction  ;  or,  in  other 
words,  the  given  dividend  must  bear  the  same  ratio  to  the 
required  quotient,  as  the  numerator  of  the  fraction  bears  to  its 
denominator.  A  unit  or  1  is  contained  in  6,  six  times  ;  %  is 
contained  in  the  same  number,  twelve  times,  and  £,  eighteen 
times  ;  and  f  ,  half  as  many  times  as  ^,  viz.  nine  times.  The 
operation  to  obtain  this  last  quotient  is  as  follows,  6x3=18, 
the  number  of  thirds  in  six  ;  and  18-^2=9. 

For  dividing  a  whole  number  by  a  fraction,  we  have  then 
the  following  rule  : 


DIVISION  OF    FRACTIONS.  139 

Multiply  the  whole  number  by  the  denominator  of  the  fraction, 
and  divide  the  product  by  the  numerator. 

Ex.  1.  Divide  9  by  J.  Operation,  9x4=36,  and  36-?-3 
=  12,  the  quotient. 

2.  Divide  15  by  f.     Operation,  15x6  =  90,  and  90^-5  = 
18,  quotient. 

Operations  of  a  similar  character  may  be  performed  by  can- 
celing. 

RULE  FOR  CANCELING. — Place  the  whole  number  above  a  hori- 
zontal line  and  invert  the  fractional  divisor;  that  is,  place  the 
denominator  above  the  line  and  the  numerator  below.  Cancel,  dfc. 

7 

3.  Divide  21  by  f.     Statement,  — :-o»     Canceled, — '—,  and 
7  x  7=49,  the  quotient  required. 

4.  Divide  42  by  f .     Statement,  ^-^.     Ans.  49. 


The  following  sums  may  be  divided  by  either  of  the  above 
rules. 

5.  Divide  18byf.  Ans.  42. 

6.  Divide  63byT93.  Ans.  91. 

7.  Divide  63  by  T9T.     Ans.  77. 

8.  Divide  42  by  f .  Ans.  47£. 

9.  Divide  66  by  f .  Ans.  99. 

10.  Divide  121  by  |.     Ans.  138f 

11.  Divide  101  by  }f     Ans.  110ft-. 

12.  Divide  32  by  j,     Ans.  40. 

That  the  scholar  may  be  enabled  to  commence  understand- 
ingly  the  division  of  fractions  by  fractions,  he  may  turn  back 
and  review  the  7th  section  of  the  introductory  remarks  of  this 
rule.  It  is  there  said,  that  a  fraction  is  divided  by  another 
fraction  by  inverting  the  divisor  and  then  multiplying  them 
together  as  in  multiplication. 

1 .  Divide  I  by  f .     Ans.  f  £  or  1 

2.  Divide  f  by  £.     Ans.  V6  or 

RULE  FOR  CANCELING. — Proceed  in  all  respects  as  in  multi- 
plication of  fractions,  in  arranging  the  terms  of  the  dividend, 


140  DIVISION  OF  FRACTIONS. 

then  invert  the   divisor  ;   that  is,  place  the  numerators   below 
and  the   denominators  above  the  line.      Proceed  to  cancel,  fyc. 

Ex.  3.  Divide  1|  by  A. 

3     3 

Statement,  ^?      Canceled,  ^|. 
10.     o  i-tx.     « 

4 

3x3  =  9,  for  2£,  Ans. 

Note.  —  When  the  divisor  is  a  compound  fraction,  each  frac- 
tion in  the  divisor  must  be  inverted. 

*>      j       j      c 

4.  Divide  f  of  £  by  ^  of  J.    Statement,  '  '    '    '    .     Canceled, 

'    '  ~^—  ;    therefore,  |  or  4  is  the  answer  required. 
*.  a.   J.  a 

5.  Divide  if  of  H  by  f  off     Statement,  l*i_l*i.i. 


or 


6.  Divide  fof  T6T  of  |  by  i  of  f  of  ^.     An^.  |,  or 

7.  Divide  ^  by  i.     ^in^.  2. 


8.  Divide  ^  of  f  of  |f  of  |f  of  ff  of  i  by  £  of  $  of  • 

of  A-     ^^•fyf-1T^5- 

9.  Divide  4f  by  ^  of  6.     Ans.  l-^. 

10.  Divide  6f  by  f  of  4.     Ans.  2^. 

11.  Divide  £  of  f  by  -J  of  6^.     Ans.  -£f%. 

12.  Divide  6  by  £.     Ans.  18. 

13.  Divide  3^  by  2|.     Ans.  l£. 

14.  Divide  |  by  12.     Ans.  Jg-. 

15.  Divide  £  of  fbyfoff     Ans.  Iff. 

16.  Divide  $  of  72  by  $  of  56.     Ans.  7f . 

17.  Divide  |f  by  6.     Ans. 

18.  Divide  f  by  7.     Ans. 

19.  Divide  T9^  by  3.     Ans.  T%. 

20.  Divide  ^  by  8.     Ans.  -f^. 

21.  Divide  8  by  i     Ans.  16. 

22.  Divide  12  by  |.     Ans.  18. 

23.  Divide  41  by  f .     Ans.  l^-. 

24.  Divide  35  by  f.     Ans.  49. 

25.  Bought  8  Ib.  of  coffee  for  |-^  of  a  dollar  ;   what  was  the 
cost  of  one  pound  ?     Ans.  -f^-  or  ^  of  a  dollar. 

26.  Bought  9  pounds  of  sugar  for  }£  of  a  dollar ;  what  was 
the  price  of  a  pound  ?     Ans.  j^g-  of  a  dollar. 


DIVISION    OF    FRACTIONS.  141 

27.  In  8£  weeks  a  family  consumes  84|  pounds  of  butter  ; 
how  much  is  that  per  week  1  Ans.  10^. 

QUESTIONS. — What  is  a  fraction  1  If  a  unit  be  divided  into  two 
equal  parts,  what  is  each  of  these  parts  called  1  How  is  it  written  7 
If  it  be  divided  into  three  equal  parts,  what  is  each  part  called,  and  how 
is  it  written  1  Similar  questions  should  be  asked  respecting  other  frac- 
tions. When  more  parts  than  one  are  to  be  expressed,  how  is  it  done  1 
What  do  fractions  therefore  express'?  How  are  they  represented! 
What  is  the  number  below  the  line  called  1  What  does  the  denomina- 
tor show  7  What  is  the  number  above  the  line  called]  What  does 
the  numerator  show  1  What  are  the  two  numbers  called,  when  spoken 
of  collectively  ?  How  many  kinds  of  fractions  are  there  1  What  are 
they'?  What  is  a  proper  fraction  7  Give  an  example.  What  is 
an  improper  fraction  1  Give  an  example.  What  is  a  simple  fraction  1 
Give  an  example.  What  is  a  compound  fraction  1  Give  an  example. 
What  is  a  mixed  number  1  Give  an  example.  What  is  a  complex 
fraction  1  Give  an  example.  What  does  the  denominator  show"? 
If  the  denominator  remain  the  same,  how  is  the  value  of  the  fraction 
affected  by  increasing  the  numerator  1  How,  by  diminishing  if?  Give 
an  illustration.  What  is  therefore  the  value  of  a  fraction'?  If  the 
numerator  of  a  fraction  be  less  than  the  denominator,  how  is  its  value 
compared  with  a  unit  1  If  the  numerator  be  equal  to  the  denominator, 
how  then  does  its  value  compare  with  a  unit  1  And  how,  if  the  nume- 
rator be  greater  than  the  denominator  1  What  is  the  only  consideration 
which  limits  the  value  of  a  fraction  1  In  what  ratio  is  the  value  of  a 
fraction  increased  1  How  then  may  fractions  be  multiplied  ?  How 
is  a  fraction  multiplied  into  a  number  equal  to  its  denominator?  In 
what  form  should  the  terms  of  the  fraction  always  be  preserved  ?  What 
is  necessary  to  increase  or  diminish  the  value  of  a  fraction  1  How  then 
may  a  fraction  be  multiplied  by  a  whole  number  1  In  what  two  ways 
may  fractions  be  multiplied  by  whole  numbers'?  How  may  a  frac- 
tion be  divided  by  a  whole  number  1  How  else  may  a  fraction  be 
divided  by  a  whole  number  7  In  what  two  ways  then  may  fractions  be 
divided  by  whole  numbers  1  In  what  case  should  the  numerator  always 
be  divided  1  What  operation  is  therefore  performed  on  the  value  of  a 
fraction,  when  the  numerator  is  operated  upon  ?  And  what,  when  the 
denominator  is  operated  upon  1  How  is  a  fraction  multiplied  by  a  frac- 
tion 1  How  is  a  fraction  divided  by  a  fraction?  If  the  numerator  and 
denominator  be  both  multiplied  by  the  same  number,  how  is  the  value 
of  the  fraction  affected  1  What  is  the  rule  for  reducing  fractions  to 
their  lowest  terms  1 

What  is  the  rule  for  reducing  a  whole  number  or  mixed  quantity  to 
an  improper  fraction  ?  How  is  an  improper  fraction  reduced  to  a  whole 
or  mixed  number?  In  all  cases  of  division,  what  disposition  may  be 
made  of  the  remainder,  when  any  occurs  ?  How  are  compound  frac- 
tions reduced  to  simple  ones  ?  What  is  the  rule  for  canceling  ?  What 
is  note  1st  ?  What  is  note  2d  ?  What  is  note  3d  ?  What  is  Case  5th  ? 
How  arc  fractions  of  low  denominations  reduced  to  those  of  higher 
denominations  ?  What  is  the  rule  for  canceling  ? 

How  are  fractions  of  high  denominations  reduced  to  those  of  a  lower 
value  ?  What  is  Case  6th  ?  What  is  the  rule  for  it  ?  What  is  Case 
7th  1  What  is  the  rule  for  it  ?  What  is  the  rule  for  canceling  ? 


142  DECIMAL  FRACTIONS. 

What  is  Case  8th  1  What  is  the  rule  for  if?  What  is  the  note  1 
What  must  be  done  before  fractions  can  be  added  1  What  else  requi  res 
to  be  done!  What  is  the  rule  for  the  addition  of  fractions  1  What 
note  follows!  What  is  note  2d7  What  preparations  are  necessary 
before  fractions  can  be  subtracted  1  What  is  the  rule  for  subtracting 
fractions  1  How  is  a  fraction  multiplied  by  a  whole  number  1  How 
is  a  fraction  multiplied  into  a  quantity  equal  to  its  denominator. 
What  is  the  rule  for  multiplying  fractions  by  fractions  1  What  is  the 
rule  for  canceling  1  Into  what  three  kinds  is  division  of  fractions 
naturally  divided  1  How  is  a  fraction  divided  by  a  whole  number  1 
What  is  the  rule  for  dividing  a  whole  number  by  a  fraction  1 
How  are  fractions  divided  by  fractions'?  What  is  the  rule  for 
canceling "?  What  note  follows  the  rule  1 


DECIMAL   FRACTIONS. 

In  the  preceding  rule  we  have  contemplated  the  unit  as  divi- 
ded into  any  number  of  equal  parts. 

We  are  now  to  regard  it  as  divided  frst  into  ten  equal  parts, 
then  each  of  these  into  ten  other  equal  parts,  or  the  whole  unit  into 
one  hundred  equal  parts  ;  and  these  parts  again,  each  into  ten 
other  parts ;  or  the  whole  into  a  thousand  equal  parts,  &c. 
The  expressions  obtained  by  these  several  divisions,  therefore, 
decrease  in  value  in  the  constant  ratio  of  ten,  from  the  left  to 
the  right,  and  are  called  decimals.  Whole  numbers,  as  was 
shown  in  Numeration,  increase  in  the  same  ratio  from  the  right 
to  the  left,  and  both  commence  their  enumeration  with  the  unit 
figure.  The  connection  between  them  is  therefore  so  intimate 
as  to  render  them  susceptible  of  being  written  together  and 
subjected  to  the  same  operations.  The  only  important  con- 
sideration in  writing  them,  in  addition  to  what  has  already 
been  explained,  is  to  distinguish  the  one  from  the  other.  This 
is  effected  by  the  period,  called  in  decimals,  the  point  of  sepa- 
ration, which  is  always  placed  between  them.  In  the  ex- 
pression, 23.56,  the  23  is  the  whole  number,  and  the  .56 
the  decimal. 

It  will  be  observed  that  decimals,  although  they  express 
parts  of  units,  do  not,  like  vulgar  fractions,  require  two  terms 
to  express  them.  The  given  decimal  may  however  be  regarded, 
as  in  truth  it  is,  a  numerator,  with  a  denominator  always  under* 


DECIMAL    FRACTIONS.  143 

stood.  What  this  denominator  is,  it  shall  be  our  next  object 
to  illustrate.  The  scholar  may  therefore,  in  the  first  place, 
carefully  examine  the  following  table  of  whole  numbers  and 
decimals.  The  decimals,  it  will  be  observed,  are  read  or 
numerated  from  the  left  to  the  right. 


a  S 
876543  2. 3  45678 

By  an  examination  of  the  preceding  table,  the  scholar  will 
see  that  the  3  on  the  right  of  the  separatrix  is  so  many  tenths. 
In  the  preceding  rule,  this  would  be  thus  expressed,  T3Q,  which, 
by  section  8th  of  the  introductory  remarks  of  the  same 
rule,  equals  T3^.  He  will  also  see,  that  the  4  on  the 
same  side  of  the  separatrix,  is  so  many  hundredths,  or  T^. 
Now  it  is  obvious,  that  these  two  fractions  united,  would  make 
"liny  The  same  process  of  reasoning  will  show,  that  the  next 
figure,  or  5,  is  so  many  thousandths  ;  and  since  T^=I\%,  if 
the  5  be  added,  the  amount  will  be  f^4^.  From  the  preceding, 
we  therefore  learn,  that  if  the  decimal  consist  of  one  figure 
only,  it  is  so  many  tenths  ;  if  it  consist  of  two  figures,  it  is  so 
many  hundredths,  and  if  of  three,  it  is  so  many  thousandths  ; 
and  from  this  we  derive  the  following  conclusion,  viz.  that  the 
denominator  of  a  decimal  fraction  always  consists  of  a  figure  1, 
with  as  many  cyphers  annexed  to  it,  as  there  are  figures  in  the 
given  decimal. 

The  scholar  may  write  a  denominator  to  each  of  the  follow- 
ing decimals,  viz.  .6  ;  .356  ;  .26  ;  .7426  ;  .98654  ;  .71639. 

From  the  preceding  explanation  of  the  nature  of  decimals, 
it  is  obvious,  that  cyphers  added  to  the  right  of  a  decimal 
do  not  effect  its  value,  while  those  placed  on  the  left  dimin- 
ish its  value  in  a  ten-fold  ratio.  For  .5  is  the  same  in 
value  as  T5^,  or  \.  Now  if  a  cypher  be  added  to  the  right  of 
the  .5,  it  becomes  .50,  which  is  of  equal  value  with  -j^,  and 
this  also  equals  £.  (See  Case  1st,  Vulgar  Fractions. )  If, 
then,  .5  and  .50  are  each  equal  to  2,  it  is  obvious,  that  the 
value  of  a  decimal  is  not  affected  by  cyphers  placed  on  the 


144  DECIMAL  FRACTIONS. 

right  of  it.  But  if  they  be  placed  on  the  left,  they  diminish 
the  value  of  the  decimal  in  a  ten-fold  ratio.  The  decimals  .5, 
.05,  and  .005,  will  serve  as  an  illustration.  From  the  explana- 
tion given  above,  the  denominators  of  these  several  decimals 
are  10,  100,  and  1000  ;  or  the  decimals  may  be  thus  written  : 
To*  nn>»  TOlV  But  ^ve  hundredths  equal  only  one  tenth  part 
of  five  tenths  ;  and  five  thousandths,  one  tenth  part  of  five 
hundredths.  Hence,  cyphers  placed  on  the  left  of  a  decimal 
diminish  its  value  as  above  specified. 

The  following  numbers  may  now  be  expressed  by  figures, 
and  then  read. 

1.  Seventy-six  and  six  tenths.     Ans.  76,6. 

2.  One  and  three  hundredths.     Ans.   1.03. 

3.  Eighty  and  fifty-eight  thousandths.      80.058. 

4.  One  hundred  and  fifty-six,  and  thirty-nine   thousandths. 
Ans.   156.039. 

5.  One   hundred    and   one,   and   five   thousandths.      Ans. 
101.005. 

The  scholar  will  observe,  that  if  there  is  but  one  decimal 
figure,  and  that  tenths,  it  requires  the  point  only  to  be  placed 
at  the  left  of  it,  to  express  its  true  value  ;  if  it  be  hundredths, 
it  requires  a  cypher  to  be  placed  at  the  left  of  it,  and  if  it  be 
thousandths,  it  requires  two  cyphers  to  be  thus  placed,  with 
the  decimal  point  on  the  left  of  the  cyphers  ;  and  so  on  accord- 
ing to  the  denomination. 

6.  Write  down  nine,  and  three  hundred  thousandths.     Ans. 
9.00003. 

7.  Write  down  twelve  and  one  millionth.    Ans.  12.000001. 

8.  Three  hundred  and  seventy-five,  and  seven  tens  of  thou- 
sandths.  Ans.   375.0007. 

9.  Ninety-five  hundredths. 

10.  Three  hundred  and  sixteen  thousandths. 

11.  Forty-five  millionths. 

12.  Sixty-nine,  and  nine  hundred  and  three  thousandths. 

13.  Four  hundred  and  fifty-six,  and  seventeen  millionths. 

14.  Five  hundred,  and  three  tens  of  millionths. 

15.  One,  and  six  hundred  of  millionths. 

16.  Eleven,  and  seven  billionths. 

17.  Seven  hundred  and  sixty -two  billionths. 

1 8.  Four  hundred  and  twenty-one,  and  nineteen  thousandths . 


DECIMAL  FRACTIONS.  145 

19.  Seven  hundred  and  six,  and  one  hundred  and  three 
millionths. 

20.  Twelve  hundred  and  six  trillionths. 

The  scholar  is  now  requested  to  turn  back  to  Federal  Money, 
and  compare  the  denominations  there  given,  with  those  here 
brought  to  view.  He  will  there  find  the  dollar  given  as  the 
unit  money ;  the  dime  as  the  tenth  part  of  the  dollar ;  the 
cent  as  the  tenth  part  of  the  dime,  or  the  hundredth  part  of  the 
dollar  ;  and  the  mill  as  the  tenth  part  of  the  cent,  the  hundredth 
part  of  the  dime,  and  the  thousandth  part  of  the  dollar.  It  is 
therefore  obvious,  that  Federal  Money  and  decimals  are  ope- 
rated upon  by  the  same  general  principles. 

1.  Reduce  $21,  8  dimes,  and  6  mills,  to  mills.    Ans.  21806. 

2.  Reduce  21806  mills  to  dollars,  cents,  and  mills.     Ans. 
$21.806. 

3.  Reduce  $12,  3  dimes,  4  cents,  and  9  mills,  to  mills.  Ans. 
12349. 

4.  Reduce  12349  mills  to  dollars,  cents,  and  mills.     Ans. 
$12.349. 

5.  Reduce  $25  to  cents.     Ans.  2500. 

6.  Reduce  $9  to  mills.     Ans.  9000. 

To  reduce  dollars  to  cents,  we  therefore  add  two  cyphers, 
and  to  reduce  them  to  mills,  we  add  three. 

7.  Reduce  2567  cents  to  dollars.     Ans.   $25.67,  or   $25 
and  67  cents. 

8.  Reduce  38679  mills  to  dollars,  &c.     Ans.  $38.679,  or 
$38,  67  cents,  and  9  mills. 

To  reduce  cents  to  dollars,  we  therefore  cut  off  two  figures, 
and  to  reduce  mills  to  the  same,  we  cut  off  three  figures  from 
the  right  of  the  given  number. 

9.  Reduce  $2  to  mills. 

10.  Reduce  99  cents  to  mills. 

11.  Reduce  $1.03  to  mills. 

12.  Reduce  467  cents  to  dollars. 

13.  Reduce  12008  mills  to  dollars. 

14.  Reduce  $42  and  3  mills  to  mills. 

15.  Reduce  9000  mills  to  dollars. 

13 


146  ADDITION  OF  DECIMAL  FRACTIONS. 


ADDITION  OF   DECIMAL  FRACTIONS. 

The  scholar  must  here  exercise  a  good  degree  of  caution 
in  writing  the  numbers  to  be  added.  He  will  recollect  that, 
in  adding  vulgar  fractions,  it  was  necessary  to  reduce  them  all 
to  the  same  name  or  denominator,  before  the  numerators  could 
be  added ;  and  that  in  Simple  and  Compound  Addition,  the 
same  denominations  only  could  be  united.  The  same  is  true 
of  Decimal  Fractions.  Hence,  we  have  the  following  rule  : 

RULE. — Place  the  whole  numbers  as  in  Simple  Addition, 
with  units  under  units,  and  tens  under  tens,  §c.  A  Iso  place 
the  decimals  on  the  right  of  the  whole  numbers,  with  tenths  under 
tenths,  hundredths  under  hundredths,  and  thousandths  under 
thousandths,  fyc. ;  then,  beginning  with  the  lowest  denomination, 
add  up  and  carry  as  in  Simple  Addition.  Lastly,  from  the 
amount,  point  of  as  many  decimals  as  are  equal  to  the  greatest 
number  of  decimals  in  any  one  of  the  given  numbers. 

Ex.1.  What  is  the  amount  of  3.56;  42.923;  125.6;  4.32; 
and  59.365. 

ADDED. 

3.56 
42.92  3 
12  5.6 

4.32 
59.365 


2  3  5.7  6  8 

The  greatest  number  of  decimals  in  either  of  the  numbers 
given,  is  three  ;  therefore,  three  decimals  are  to  be  cut  off  from 
the  sum.  It  will  always  be  found  correct  to  place  the  decimal 
point  in  the  amount  directly  below  those  of  the  given  num- 
bers. 

2.  Add  the  following  numbers,   325.63;  275.215;  1.02; 
17.653  ;  136.1.     Amount,  755.618. 

3.  What  is  the  amount  of  72.5;  32.071;  2.1574;  371.4; 
2.75  ?     Ans.  480.8784. 


, 


SUBTRACTION  OF  DECIMALS.  147 

4.  What  is  the  amount  of  225.75  ;  25.50  ;  8.255  ;  27.225  ? 
Ans.  286.73. 

5.  What  is  the  amount  of  35.175;  75.15;   13.31;  25.755? 
Ans.  149.39. 

6.  What  is  the  amount  of  304.39;  291.09;  136.99;  12.10? 
Ans.  744.57. 

7.  What  is  the  amount  of  365.541  ;  487.06  ;  94.67  ;  472.5  ; 
439.089?     Ans.  1858.860. 

8.  What  is  the  amount  of  2.151  ;  375.422;  .675;  .4567? 
Ans.  378.7047. 

9.  Add  together  the  following  decimals,  viz.  sixteen  hun- 
dredths  ;  two  hundred  and  thirty-five  thousandths  ;  six  tenths  ; 
and  one  thousandth.     Ans.  .996. 

10.  What  is  the  sum  of  one  tenth;  two  hundredths  ;  three 
thousandths  ;    four   tens    of   thousandths  ;    five   hundreds   of 
thousandths,  and  six  millionths  ?     Ans.  .123456. 

11.  WThat   is  the   amount  of  $72.375;  $41.176;  $1.47; 
$395.20  ;  $56.65  ;  $146.73,  and  $0.16.     Ans.  $713.761. 

12.  Bought  a  yoke  of  oxen  for  $121.56  ;  a  horse  for  $156.- 
375  ;  a  cow  for  $37.086  ;  and  a  quantity  of  grain  for  $95.739. 
What  was  the  cost  of  the  whole  ?     Ans.  $410.760. 


SUBTRACTION   OF    DECIMALS. 

The  scholar  will  need  no  explanation  of  the  nature  of  this 
rule.  His  knowledge  of  Subtraction,  and  of  the  peculiarity 
of  decimals,  will  enable  him  at  once  to  make  a  correct  appli- 
cation of  the  following  rule  : 

RULE. — Set  down  the  less  number  under  the  greater,  so  that 
each  figure  in  the  lower  number,  or  subtrahend,  shall  stand  di- 
rectly under  one  of  its  own  name  or  denomination.  Proceed 
to  subtract  as  in  simple  numbers,  and  place  the  separatrix  as  in 
addition  of  decimals. 

Ex.  1.  From  378.635  take  195.275. 


148  MULTIPLICATION  OF  DECIMALS. 

OPERATION. 

378.635 
1  95.2  75 


1  83.360  remainder. 

2.  From  462.3  take  218.15.     Rem.  244.15. 

3.  From  16.705  take  7.6845.     Rem.  9.0205. 

4.  From  132.4  take  36.36.     Rem.  96.04. 

5.  From  127.05  take  66.006.     Rem.  61.044. 

6.  From  100.001  take  77.77.     Rem.  22.231. 

7.  From  five  hundred  and  thirty-six,  and  fifteen  hundredths, 
take  two  hundred   and  thirty-six,  and  eighteen  hundredths. 
Rem.  299.97. 

8.  From  six  dollars  take  fifty-five  cents.     Rem.  $5.45. 

9.  From  one  dollar  take  one  mill.     Rem.  .999  mills. 

10.  From  16  dollars  take  nineteen  cents  and  one  mill.    Rem. 
$15.809. 


MULTIPLICATION    OF    DECIMALS. 

RULE. — Multiply  as  in  whole  numbers,  and  point  off  as 
many  decimals  in  the  product  as  there  are  decimals  in  both  fac- 
tors. Whenever  the  decimals  in  the  product  are  not  as  many 
as  those  of  the  factors,  the  deficiency  must  be  supplied  by  pla- 
cing cyphers  on  the  left  of  them. 

Ex.  1.  Multiply  25.16  by  3.45. 

25.16 
3.45 


12580 
10064 

7548 

8  6.8  0  2  0 
Four  figures  are  cut  off  in  the  product  as  decimals,  in  ac- 


DIVISION  OF  DECIMALS.  149 

V 

cordance  with  the  rule,  there  being  four  decimals  in  the  two 
factors. 

2.  Multiply  175.2  by  45.72.     Ans.  8010.144. 

3.  Multiply  15.75  by  1.05.     Ans.  16.5375. 

4.  Multiply  37.99  by  25.77.     Ans.  979.0023. 

5.  Multiply  100.00  by  0.01.     Ans.  1.0000. 

6.  Multiply  3.45  by  .16.     Ans.  .5520. 

7.  Multiply  25.238  by  12.17.     Ans.  307.14646. 

8.  Multiply  27.56  by  12.22.     Ans.  336.7832. 

9.  Multiply  .01  by  .01.     Ans.  .0001. 

10.  Multiply  7.02  by  5.27.     Ans.  36.9954. 

11.  Multiply  .001  by  .001.     Ans.  .000001. 

12.  Multiply].25  cents  by  .25  cents.     Ans.  .0625,  or  6£  cts. 

Note. — To  multiply  a  decimal  by  10,  100,  1000,  &c.,  it  is 
necessary  only  to  remove  the  decimal  point  as  many  places  to 
the  right  as  there  are  cyphers  in  the  multiplier. 

13.  Multiply  1.56  by  10.     Ans.  15.6. 

14.  Multiply  36.541  by  100.     Ans.  3654.1. 

15.  Multiply  .42  by  100.     Ans.  42. 

16.  Multiply  46.3789  by  1000.     Ans.  46378.9  . 


DIVISION   OF   DECIMALS. 

We  are  now  to  reverse,  the  preceding  operation.  In  mul- 
tiplying decimals,  we  were  directed  to  point  off  as  many  deci- 
mal figures  in  the  product  as  there  were  in  both  factors.  In 
Division,  the  dividend  corresponds  to  the  product  in  Multipli- 
cation, and  the  divisor  to  one  of  the  factors  which  produced 
that  dividend,  and  we  are  required  to  obtain  the  other  factor. 
Therefore  the  decimals  of  the  quotient  and  divisor  united  must 
equal  those  of  the  dividend. 

RULE. — Divide  as  in    Simple  Numbers,  and  point  off  from 
the  right  of  the  quotient  as  many  decimals  as  are  equal  to 
13* 


150  DIVISION  OF  DECIMALS, 

the  excess  of  decimals  in  the  dividend,  over  those  in  the  di- 
visor. 

Note  1st. — If  the  decimal  places  in  the  divisor  be  more 
than  those  in  the  dividend,  annex  cyphers  to  make  them 
equal. 

2d.  If,  after  dividing,  there  be  a  remainder,  cyphers  may 
be  annexed  to  the  remainder  and  the  division  continued.  The 
cyphers  thus  added  are  decimals. 

3d.  If  the  decimals  in  the  divisor  and  dividend  are  equal, 
and  there  is  no  remainder  after  dividing,  the  quotient  will  be  a 
whole  number. 

4th.  If  the  figures  in  the  quotient  do  not  equal  the  excess 
of  decimal  places  in  the  dividend  over  those  of  the  divisor, 
supply  the  defect  by  prefixing  cyphers. 

5th.  To  divide  the  decimal  number  by  10,  100, 1000,  &c., 
it  is  necessary  only  to  remove  the  point  as  many  figures  to  the 
left,  as  there  are  cyphers  in  the  divisor. 

Ex.  1.  Divide  34.317  by  21.75. 

PERFORMED. 

2  1.7  5)  3  4.3  17(1.577  + 
21.75 


12567 
10875 

16920 
15225 

]  6  950 
15225 


1725  remainder. 

In  the  solution  of  this  example,  two  cyphers  have  been  ad- 
ded  to  the  remainders  of  the  dividend.  By  Note  2d,  the 
whole  number  of  decimals  in  the  dividend  is  therefore  five, 
and  there  are  two  in  the  divisor  ;  three  should,  therefore,  be 


DIVISION  OF  DECIMALS.  151 

cut  off  from  the  quotient.     The  plus  sign  in  the  quotient  al- 
ways implies  a  remainder. 

2.  Divide  30515.50  by  100.     Ans.  305.1550. 

For  the  solution  of  the  preceding  sum,  see  Note  5th. 

3.  Divide  483.125  by  386.5.     Ans.  1.25. 

4.  Divide  198.15625  by  186.5.     Ans.  1.0625. 

5.  Divide  .56  by  1.12.     Ans.  .5. 

6.  Divide  99.99  by  33.3.     Ans.  3.0027. -f 

7.  Divide  1.00  by  .12.     Ans.  8.333.+ 

8.  Divide  14325.16  by  1.33.     Ans.  10770.721.  -4- 

9.  Divide  36.5  by  10.     Ans.  3.65. 

10.  Divide  36.5  by  100.     Ans.  .365. 

11.  Divide  981  by  1000.     Ans.  .981. 

12.  Divide  543.67  by  3.46.    Ans.  157.13.+ 

APPLICATION. 

1.  If  36.34  bushels  of  corn  grow  on  an  acre,  how  many 
acres  will  produce  674  bushels  1     Ans.  17.804  acres. 

2.  If  6  yards  of\cloth  cost  $24.48,  what  was  the  price  per 
%yard?     Ans.  $4.08. 

3.  Bought  56.87  yards  of  cloth  at  $2.31  per  yard ;  what  was 
the  whole  cost  ?     Ans.  $131.3697. 

4.  The  first  of  three  men  possessed  $685.423  ;  the  second, 
$746.03  ;  and  the  third,  $10864.273.     How  much  had  they 
all?     Ans.  12295.726. 

5.  What  cost  9.6  yards  of  cloth  at  $6.42  per  yard  ?     Ans. 
$61.632. 

6.  If  a  man  earn   2  dollars  2  mills  per  day,  how  much 
would  he  earn  in  93.5  days  ?     Ans.  $187.1870. 

7.  What  cost  .675  of  a  cord  of  wood  at  $3  a  cord  ?     Ans. 
2.025. 

8.  If  a  yard  of  cloth  cost  $5.5625,  how  much  will  .25  of  a 
yard  cost?     Ans.  1.3906. 


152  REDUCTION  OF  FRACTIONS. 


REDUCTION    OF    VULGAR   AND 
DECIMAL    FRACTIONS. 

The  value  of  a  vulgar  fraction  is  the  quotient  arising  from 
dividing  the  numerator  by  the  denominator.  Therefore, 

CASE  1st. — To  REDUCE  A  VULGAR  FRACTION  TO  A  DECIMAL. 

RULE. — Annex  cyphers  to  the  numerator,  and  divide  it  by 
the  denominator. 

Note. — If  the  given  fraction  be  proper,  the  quotient  will 
always  be  a  decimal,  and  will  consist  of  figures  equal  in  num- 
ber to  the  cyphers  annexed ;  or,  if  the  number  of  figures  be  less, 
cyphers  must  be  prefixed  to  complete  the  number. 


Ex.  1 .  Reduce  £  to  a  decimal. 

A.T 

(,0 
.  7  5  the  decimal  required. 


OPERATION. 
4)3.00 


2.  Reduce  J  to  a  decimal. 

OPERATION. 
5  )1.0 

.  2  the  decimal  required. 

3.  Reduce  |  to  a  decimal.     Ans.  .333.+ 

4.  Reduce  |- to  a  decimal.     Ans.  .125. 

5.  Reduce  if  to  a  decimal.     Ans.  .9375. 

6.  Reduce  ^  to  a  decimal.  ^4ns.  .1. 

7.  Reduce  rf  to  a  decimal.      Ans.  .923.+ 

8.  Reduce  -f^  to  a  decimal.     Ans.  .9. 

9.  Reduce  f  to  a  decimal.     Ans.   .666.+ 

'CASE  2d. — To  REDUCE  A  DECIMAL  TO  A  VULGAR  FRACTION. 

RULE. — Write  down  the  given  decimal  as  a  numerator,  and 
for  a  denominator,  write  1  with  as  many  cyphers  annexed  as 
there  are  figures  in  the  numerator,  and  then  reduce  the  fraction 
to  its  lowest  terms.  (See  remarks  introductory  to  Decimals.) 


REDUCTION  OF  FRACTIONS.  153 

1.  Reduce  the  decimal  .125  to  a  vulgar  fraction.     Per- 
formed,  xo2^'    (see   Case   lst'   Vulgar   Fractions,) 

»  and  again,  ^-^25=4,  Ans. 

2.  Reduce  .75  to  a  vulgar  fraction.     Performed, 
=  f,  Ans. 

3.  Reduce  .9385  to  a  vulgar  fraction.     Ans. 

4.  Reduce  .2  to  a  vulgar  fraction. 

5.  Reduce  .16  to  a  vulgar  fraction. 

6.  Reduce  .25  to  a  vulgar  fraction. 

7.  Reduce  .45  to  a  vulgar  fraction. 

8.  Reduce  .55  to  a  vulgar  fraction. 

9.  Reduce  .8  to  a  vulgar  fraction. 

10.  Reduce  .24  to  a  vulgar  fraction. 

11.  Reduce  .945  to  a  vulgar  fraction. 

12.  Reduce  .844  to  a  vulgar  fraction. 

CASE  3d. — To  REDUCE  LOWER  DENOMINATIONS  TO  DECIMALS 

OF  A  HIGHER. 

RULE  1st. — Write  down  the  several  denominations  which  are 
to  be  reduced  to  decimals  of  a  higher  denomination,  one  above 
another,  with  the  lowest  uppermost ;  then  divide  each  denomina- 
tion^ commencing  with  the  lowest,  by  that  number  which  is  re- 
quired of  each  to  make  a  unit  of  the  next  higher  denomination, 
and  at  each  division  place  the  quotient  as  a  decimal  on  the  right 
of  the  next  higher  denomination.  The  number  last  obtained 
will  be  the  required  decimal. 

Note. — It  will  be  obvious  that  the  division  of  the  lowest  de- 
nomination must  be  effected  by  adding  cyphers  to  that  denomi- 
nation. Cyphers  must  also  be  added  to  each  higher  denomi- 
nation to  reduce  them,  unless  the  decimal  figures  previously 
obtained  be  sufficient. 

The  reason  of  the  above  rule  is  readily  shown.  Suppose  it 
is  required  to  reduce  7  pence  to  the  fraction  of  a  shilling.  The 
fraction  would  be  T7^,  because  the  shilling  is  divided  into  12 
equal  parts,  and  7  of  these  parts  are  taken,  and  this  vulgar  frac- 
tion is  reduced  to  a  decimal  by  adding  cyphers  to  its  numera- 
tor and  dividing  by  its  denominator.  (See  Case  1st  of  Deci- 
mals. ) 


154  REDUCTION  OF  FRACTIONS. 

Ex.  1.  Reduce  7  s.  6  d.  to  the  decimal  of  a  pound  sterling. 
The  statement  would  be,  ^  ^.     That  is,  the  6  d.  is  to  be  divi- 
ded by  12,  and  the  7  s.  by  20.     This  must  be  effected  by 
adding  cyphers. 

OPERATION. 


12 
20 


6.0 
7.5  0  0 


.3  7  5= required  decimal. 
2.  Reduce  9  d.  2  qr.  to  the  decimal  of  a  pound  sterling. 


12 
20 


OPERATION. 
2.0 


9.5  0  0  0  0  0 


.791666 


Ans.   .0395833  + 

RULE  2d. — Reduce  the  given  quantity  to  its  lowest  denomi- 
nation, and  divide  it  by  a  unit  of  the  denomination  of  the  re- 
quired fraction,  reduced  to  the  same  denomination. 

Ex.  3.  Reduce  10s.  9d.  2qr.  to  the  decimal  of  a  pound 
sterling.  10s.  9  d.  2qr.  —  518  qr. ;  and  by  the  rule,  518  qr. 
are  to  be  divided  by  1  £.  reduced  to  qr.  viz.  [by  960  qr. 
Therefore,  |i|  is  the  fractional  answer,  and  518-^960= 
.539583  +  ,  Ans. 

To  understand  the  above  operation  the  scholar  should  re- 
member that  10s.  9  d.  2  qr.  or  518  qr.  are  to  be  divided  into 
as  many  equal  parts  as  there  are  farthings  in  1  £.  =  960  qr.,  and 
one  of  these  parts  =  §-J§,  or  the  decimal  .539583.+ 

4.  Reduce  9  s.  8  d.  to  the  decimal  of  a  pound  sterling.     Ans. 
.4833.+ 

5.  Reduce   3qr.   16  Ib.  to  the  decimal  of  a  cwt.  ?     Ans. 
.8928571.+ 

6.  Reduce  16  £.  12s.  8  d.  to  a  decimal  expression.     Ans. 
16.633333.+ 

7.  Reduce  3qr.  2na.  to  the  decimal  of  a  yard.     Ans.  .875. 

8.  Reduce  2  roods  and  20  rods  to  the  decimal  of  an  acre. 
Ans.  .625. 


REDUCTION  OF  FRACTIONS.  155 

9.  Reduce  3  furlongs  16  rods  to  the  decimal  of  a  mile. 
Ans.  .425. 

10.  Reduce  12  hours,  15  minutes,  and  30  seconds  to  the  deci- 
mal of  a  day.     Ans.  .51076.+ 

1 1 .  Reduce  2  cwt.  3  qr.  24  Ib.  to  the  decimal  of  a  ton.     Ans. 
.14821428. 

CASE  4th. — To  FIND  THE  VALUE  OF  A  DECIMAL  IN  INTEGERS 

OF  LOWER  DENOMINATIONS. 

RULE. — Multiply  the  decimal  by  the  number  required  to  re- 
duce it  to  the  next  lower  denomination,  and  from  the  right 
hand  of  the  product  cut  off  as  many  figures  as  there  are  in  the 
given  decimal.  The  figures  on  the  left  of  the  point  will  be 
integers  of  the  denomination  next  below  that  given.  Proceed 
in  the  same  way  through  all  the  denominations,  and  the  figures 
on  the  left  of  the  several  points  will  be  the  answer  required. 

This  rule  being  directly  the  reverse  of  the  preceding,  needs 
no  explanation. 

Ex.  1.  What  is  the  value  of  the  decimal  .5638  of  a  pound 
sterling  ? 

OPERATION. 

.5638 

2  0 


1  1.276  0=the  shillings  and  decimals  of  a  shilling 
12         in  .5638  of  a  pound  sterling. 

3.312  O^zthe  pence  and  decimals  of  a  penny  in 
4         .2760  of  a  shilling. 

1.2  4  8  0=the  farthings  and  decimals  of  a  far- 
thing in  .3120  of  a  penny. 

The  value  of  the  above  decimal  in  shillings,  pence,  &c.  is 
11s.  3d.  1.2480qr. 

Note.— The  integers  on  the  left  of  the  points  are  the  num- 
bers which  compose  the  answer. 

2.  What  is  the  value  of  .75  of  a  pound  sterling?     Ans.  15s. 

3.  What  is  the  value  of  .53854  of  a  pound  sterling  ?     Ans. 
10s.  9  d.  1  qr.  nearly. 


156  REDUCTION  OF  FRACTIONS. 

4.  What  is  the  value  of  .625  of  an  acre  1     Ans.  2  roods,  20 
rods. 

5.  What  is  the  value  of  .148712678  of  a  ton  ?     Ans.  2  cwt. 
3qr.  25  Ib.  1  oz.+ 

6.  What  is  the  value  of   .676  of  a  cwt.  ?     Ans.  2  qr.  19  Ib. 
11  oz.+ 

7.  How  many  furlongs,  &c.  in  .425  of  a  mile  ?     Ans.  3  fur. 
16  rods. 

8.  How  many  quarters,  &c.  in  .66  of  a  yard  1     Ans.  2  qr. 
2  nails,  1.26  inches. 

9.  How  many  roods,  &c.  in  .321  of  an  acre  1     Ans.  I  rood, 
11  rods,  10yd.  8  feet. 

10.  What  is  the  value  of  .875  of  a  hogshead  of  wine  ?     Ans. 
55  gal.  1  pint. 

11.  What  is  the  value  of  .875  of  a  yard  ?     Ans.  3  qr.  2  nails. 

12.  What  is  the  value  of  .9   of  an  acre  1     Ans.   3  roods,  24 
rods. 

QUESTIONS. — How  have  we  contemplated  the  unit  as  divided  in  the 
preceding  rule  1  How  are  we  to  regard  it  as  divided  in  this  rule  ?  In 
what  ratio  do  the  several  divisions  in  decimals  decrease  1  What  are 
they  called  1  What  relation  is  there  between  integers  and  decimals  ? 
What  is  the  only  important  consideration  which  has  not  already  been 
explained  1  How  many  terms  are  required  to  express  decimals  7 
How  may  that  term  be  regarded  1  Repeat  the  table  of  whole  numbers 
and  decimals.  Of  what  does  the  denominator  of  decimal  fractions 
always  consist  1  How  do  cyphers  added  to  the  right  of  a  decimal, 
affect  its  value  7  Give  the  illustration.  How  do  cyphers  placed  on 
the  left  of  a  decimal  affect  its  value  1  Give  the  illustration.  How  do 
decimals  and  the  denominations  of  federal  money  correspond  1  How 
do  we  reduce  dollars  to  cents'?  How  do  we  reduce  them  to  mills  ? 
How  do  we  reduce  cents  and  mills  to  dollars  1  What  is  the  rule  for 
the  addition  of  decimals  1  What  is  it  for  subtracting  decimals  ?  What 
is  the  rule  for  multiplying  decimals  ?  How  does  division  stand  related 
to  multiplication  1  What  is  the  rule  for  division  of  decimals  1  What 
is  the  rule  for  pointing  off  decimals  in  addition  of  decimals'?  What 
in  subtraction  1  What  in  multiplication?  And  what  in  division? 
On  what  does  the  value  of  a  fraction  depend  1  What  is  the  rule  for 
reducing  vulgar  fractions  to  decimals  1  What  is  Case  2d  ?  What  is 
the  rule  for  if?  What  is  Case  3d?  What  is  the  rule1?  What  note 
follows  the  rule  1  Explain  the  nature  of  the  rule.  What  is  the 
second  rule  1  What  is  Case  4th  1  What  is  the  rule  for  it  ?  What 
note  follows  the  rule  ? 


REDUCTION  OF  CURRENCIES.  157 


REDUCTION   OF    CURRENCIES. 

The  currency  of  the  United  States  was  originally  pounds* 
shillings,  and  pence,  the  same  as  in  England.  This,  however, 
was  abolished  by  an  act  of  Congress,  in  1786,  and  Federal 
Money,  consisting  of  dollars,  cents,  and  mills,  was  adopted. 

The  following  table  gives  the  value  of  the  dollar  in  the  old 
currency  of  several  states  : 

In  the  New  England,  Virginia,  Kentucky,  and  Tennessee 
currency,  $l=6s.  and  1  £.  =  20s.  ;  therefore,  1  -£.  =  2¥°  or  L° 
of  $1  ;  or  $l=rT^of  l£. 

In  New  York  and  North  Carolina  currency,  $1—  8s.  and 
U.  =  20s.  ;  therefore,  l£.rr2F°=f  of  $1,  or  $l=f  of  1  £. 

In  New  Jersey,  Pennsylvania,  Delaware,  and  Maryland 
currency,  $1  =7  s.  6  d.=  90  d.  I  £.=20  s.  =240  d.-,  therefore, 
1  £.  =  %<»  =f  of  $L  or,  $l  =  f  of  1  JB. 

In  South  Carolina  and  Georgia  currency,  $l=4s.  8d.= 
56  d.  and  l£.=20  s.=240  d.;  therefore,  !£.  =  2-±j>  =  3T°  of  $1, 
or  $l=^j-  of  1  £. 

In  Canada  currency,  $1  =  5  s.  ;  therefore,  1  £.=^  of  $1,  or 


That  the  scholar  may  understand  why  the  dollar  is  com- 
posed of  a  different  number  of  shillings  and  pence  in  the  diffe- 
rent states,  it  is  necessary  only  to  say,  that  these  states  (or 
colonies,  as  they  were  at  first  called)  originally  issued  each 
their  own  money,  in  pounds,  shillings,  and  pence,  the  value  of 
which  soon  depreciated.  This  depreciation  was  greater  in 
some  states  than  in  others  ;  and  hence,  when  federal  money 
was  adopted,  more  of  the  old  currency  was  required  in  some 
states  than  in  others,  to  equal  the  dollar  of  the  new  curren- 
cy. The  value  of  this  dollar  was  4s.  6  d.  sterling  money, 
or  English  currency  ;  while  6  s.  New  England  currency,  8  s. 
New  York,  7s.  6  d.  New  Jersey,  and  4s.  8  d.  Georgia  cur- 
rency, were  required  to  equal  the  same  value. 

To  understand  changing  these  several  currencies  to  dollars, 
cents,  and  mills,  the  scholar  needs  to  examine  carefully  the 
14 


158  REDUCTION  OF  CURRENCIES. 

preceding  table.  He  will  there  observe  that  in  New  Eng- 
land currency  1  £.  =  ^  of  $1.  Therefore,  if  pounds  of  that 
currency  be  multiplied  by  10,  and  the  product  divided  by  3, 
they  will  be  changed  to  federal  money,  or  dollars,  cents,  and 
mills.  A  similar  explanation  is  applicable  to  other  currencies. 

Note. — If  the  given  money  consist  of  pounds,  shillings,  and 
pence,  reduce  the  shillings  and  pence  to  the  decimal  of  a 
pound,  (see  Case  3d  of  the  Reduction  of  Vulgar  and  Decimal 
Fractions,)  and  then  proceed  according  to  the  following  rule : 

RULE. — Notice  from  the  preceding  table  what  fraction  of  one 
dollar  makes  1  £.  of  the  given  currency,  and  multiply  the  given 
sum  by  that  fraction ;  that  is,  multiply  it  by  the  numerator  and 
divide  the  product  by  the  denominator. 

Ex.  1.  Reduce  72  £.  New  England  currency  to  federal 
money.  In  N.  E.  currency,  the  pound  =^  of  one  dollar; 
therefore, 

7  2 
1   0 

3)7  2  0 
$  2  4  0  Ans. 
2.  Reduce  75  £.  6  s.  8d.  N.  E.  currency  to  federal  money. 


12 
20 


8.0 
6.6666-f 

.333  3=the  decimal  value  of  6  s. 
8  d. ;  therefore, 

7  5.3  3  3  3 

1  0 


3  )7  5  3,3  3  3  0 
$251.111  Ans. 

3.  Reduce  15  £.  6  s.  6  d.  New  York  currency  to  federal 
money. 


REDUCTION  OF  CURRENCIES.  159 

12  I  6. 

20     6.5 


.3  2  5= the  decimal   value   of 
6s.  6d. ;  therefore, 

1  5.3  2  5 
5 


2  )  7  6.6  2  5 


$  3  8.3  1  2  5  Ans. 

To  abbreviate  the  given  operation  by  canceling,  the  follow-' 
ing  rule  may  be  adopted  : 

RULE  FOR  CANCELING. — Place  the  given  sum  above  a  hori- 
zontal line,  and,  noticing  as  before  what  fraction  of  a  dollar 
makes  1  £>.  of  the  given  currency,  write  its  numerator  above  the 
line,  and  its  denominator  below;  then  cancel,  multiply,  and 
divide. 

4.  Reduce  48  £.  10  s.  6  d.  New  England  currency  to  fede- 
ral money. 

12  I      6 
20     1  0.5 


.5  2  5  =  decimal  value  of  10s.  6d. 
16.175 


Therefore,  l,  and  16.175  xlO=$161.75. 

dt 

5.  Reduce  240  «6.  New  Jersey  currency  to  federal  money. 
Statement,          8 


80 


Canceled,        '      ,  and  80x8=$640,4*w. 

a. 

6.  Reduce  243  £.  New  Jersey  currency  to  federal  money. 
Ans.   $648. 

7.  Reduce    150  £.   South    Carolina   currency   to  federal 
money.     Ans.  $642.857.+ 

8.  Reduce  27  £.  New  England  currency  to  federal  money. 
Ans.  $90. 


160  REDUCTION  OF  CURRENCIES. 

9.  Reduce  304  £.  12  s.  9  d.  New  England  currency  to  fede- 
ral money.  Ans.  $1015.458. 

10.  Reduce  80  £.  New  York  currency  to  federal  money. 
Ans.  $200. 

11.  Reduce  84  £.  10  s.  6  d.  New  Jersey  currency  to  federal 
money.     Ans.  $225.40. 

12.  Reduce  100  £.  New  Jersey  currency  to  federal  money. 
Ans.  $266.666.+ 

13.  Reduce  16  £.  Canada  currency  to  federal  money.     Ans. 
$64. 

14.  Reduce  96  £.  Nova  Scotia  currency  to  federal  money. 
Ans.  $384. 

15.  Reduce    112£.    Georgia   currency  to   federal   money. 
Ans.  480. 

16.  Reduce  365  £.  10  s.  6  d.  New  York  currency  to  federal 
money.     Ans.  $913.812.+ 

17.  Reduce  29  £.  12  s.  8  d.  Canada  currency  to  federal  money. 
Ans.  $118.533.+ 

18.  Reduce  276  £.   10s.    3d.   South  Carolina  currency  to 
federal  money.     Ans.  $1185.053.+ 

19.  Reduce  45  £.   12s.   New  Jersey  currency  to  federal 
money.     Ans.  $121.60. 

20.  Reduce  42  £.  6  s.  New  England  currency  to  federal 
money.     Ans.  $141. 

The  scholar  will  now  reverse  the  preceding,  that  is,  he  will 
change  any  sum  of  federal  money  to  either  of  the  preceding 
currencies.  To  do  this  he  must  adopt  the  opposite  mode  of 
operation.  Therefore, 

RULE. — Notice  from  the  preceding  table  what  fraction  of  \  £. 
of  the  required  currency  makes  $  1 ;  then  multiply  the  given  sum 
by  the  numerator  of  that  fraction,  and  divide  the  product  by  the 
denominator. 

Ex.  1.  Reduce  $161.75  to  pounds,  shillings,  and  pence, 
New  England  currency. 

PERFORMED, 

161.75 
3 


485.2  5-^-10=48.525  £. 


REDUCTION  OF  CURRENCIES.  161 

Now  to  find  the  value  of  the  decimal  .525,  (see  Case  4th, 
Decimal  Fractions,)  .525x20^10.500  s.  and  .500x12  = 
6.000  d. ;  therefore  the  required  pounds,  shillings,  &c.  are 
48  £.  10s.  6d. 

2.  Reduce  $38.3125  to  pounds,  shillings,  and  pence,  New 
York  currency. 

38.3125      then,  .325  and  5  0  0 

2  20  12 


5)76.6250  6.500  shillings,  6.0  0  0  d. 

15.325         Therefore,  15  £.  6  s.  6  d.  is  the  answer. 

These  sums  may  also  be  solved  by  canceling.  Take  the 
following  rule  : 

RULE  FOR  CANCELING.  —  Place  the  given  dollars,  cents,  and 
mills,  above  a  horizontal  line,  and,  noticing  as  before  ivhat  frac- 
tion of  1  £.  of  the  required  currency  makes  $1,  place  the  nume- 
rator of  the  fraction  above  the  line,  and  the  denominator  below. 
Cancel,  multiply,  and  divide  for  the  answer. 

3,  Reduce  $95.75  to  pounds,  shillings,  &c.  in  New  Eng- 
land currency.     Ans.  28  £.  14s.  6  d. 

95.75.   3 
Statement,  —  -  --  . 

4.  Reduce  $648  to  pounds,  New  Jersey  currency.     Ans. 
243  £. 

fi4ft     q 

Statement,   -  —  . 


5.  Reduce  $642.857  to  pounds,  South  Carolina  currency. 
Ans.  150£. 

6.  Reduce  $578  to  pounds,  &c.  New  York  currency.     Ans. 
231  jB.  4s. 

7.  Reduce  $580  to  pounds,  &c.  Canada  currency.     Ans 
145  £. 

8.  Reduce  $742.50  to  pounds,  &c.  New  England  currency 
Ans.  222  £.  15s. 

9.  Reduce  $21.758  to  pounds,  &c.  New  England  currency 
Ans.  6£.  10s.  6d.2qr.+ 

14* 


162  SIMPLE  INTEREST. 

10.  Reduce  $141  to  pounds,  &c.  New  England  currency. 
Ans.  42  £.  6  s. 

11.  Reduce  $250  to  pounds,  &c.   Canada  currency.     Ans. 
62  £.  10s. 

12.  Reduce  $121.60  to  pounds,  &c.   New  Jersey  currency. 
Ans.  45  £.  12s. 

13.  Reduce  $475.75  to  pounds,  &c.   New  York  currency. 
Ans.  190  £.  6s. 

14.  Reduce  $89.54  to  South  Carolina  currency.     Ans.  20  £. 
17s.  10  d. 

15.  Reduce   $75  to  pounds,  &c.  New  England  currency. 
Ans.  22  £.  10s. 

16.  Reduce  $384  to  pounds,  &c.    Nova  Scotia  currency. 
Ans.  96  JB. 


.  —  What  was  the  original  currency  of  the  United  States'? 
When  this  was  abolished,  what  currency  was  substituted  1  What  are 
the  denominations  of  federal  money  1  What  is  the  value  of  the  dol- 
lar, New  England  currency  1  What  fraction  of  a  dollar  does  the 
pound  of  that  currency  equal  1  What  fraction  of  a  pound  does  the 
dollar  equal  1  Similar  questions  should  be  asked  relative  to  the  cur- 
rencies of  the  other  states.  The  scholar  may  explain  why  it  is  that 
the  dollar  is  composed  of  a  different  number  of  shillings  and  pence  in 
the  different  states.  What  note  precedes  the  rule  1  What  is  the  rule 
for  bringing  pounds,  &c.  into  dollars'?  How  are  the  terms  arranged 
for  canceling'?  What  is  the  rule  for  bringing  dollars  into  pounds, 
&c.  1  What  is  the  rule  for  canceling  1 


SIMPLE    INTEREST. 

Interest  is  an  allowance  made  for  the  use  of  money. 

It  is  computed  at  a  certain  rate  per  cent. ;  that  is,  at  a  cer- 
tain number  of  dollars  for  the  use  of  a  hundred  for  one  year. 

If  the  sum  on  interest  be  more  or  less  than  $100,  or  the 
time  during  which  it  draws  interest,  more  or  less  than  one 
year,  the  sums  paid  as  interest  must  be  in  proportion  both  to 
the  time  and  the  sum  lent. 

For  illustration:  if  $100  be  borrowed  for  one  year,  and  if 
the  interest  allowed  be  6  per  cent.,  the  interest  for  one  year 


SIMPLE  INTEREST.  163 

will  be  $6.  Now,  if  twice  that  sum,  viz.  $200,  be  borrowed 
for  the  same  time,  or  if  the  same  sum  be  borrowed  for  twice 
that  time,  viz.  for  two  years,  the  interest  will  in  either  case 
be  twice  that  sum,  viz.  $12.  .Again,  if  twice  the  sum  be 
borrowed  for  twice  the  time,  that  is,  if  $200  be  borrowed  for 
two  years,  the  sum  to  be  paid  as  interest  will  be  increased 
four-fold  ;  or  it  would  be  6  x  4 =$24. 

The  scholar  will  carefully  notice  the  following  particulars  : 

1st.  The  principal  is  the  money  lent,  or  the  sum  on  which 
interest  is  paid. 

2d.  The  interest  is  the  money  paid  for  the  use  of  the 
principal.  Legal  interest  is  that  established  by  law.  In  the 
New  England  states,  it  is  fixed  at  6  per  cent. ;  in  New  York, 
at  7,  and  in  Louisiana,  at  8  per  cent. 

3d.  The  amount  is  the  sum  obtained  by  adding  the  interest 
to  the  principal. 

4th.  The  rate  per  cent,  is  always  a  decimal  of  two  places, 
when  expressed  by  cents,  and  of  three,  when  expressed  by 
cents  and  mills. 

CASE  1st. — To  CAST  INTEREST  ON  ANY  SUM  FOR  ONE  OR  MORE 
YEARS. 

RULE. — Multiply  the  principal  by  the  rate  per  cent,  written 
as  a  decimal,  the  product  will  be  the  interest  for  one  year  ; 
which,  being  repeated  as  many  times  as  there  are  years  given, 
will  be  the  required  interest. 

Note  \st. — In  all  cases  where  no  rate  per  cent,  is  mentioned, 
six  per  cent,  is  always  implied. 

Ex.  1.  What  is  the  interest  of  $215  for  one  year,  at  6  per 
cent.? 

It  is  obvious  that  the  required  interest  will  be  6  times  215 
cents,  for  it  is  6  cents  on  each  dollar.  Therefore, 


1290  cents  =  $12.90,  Ans. 

On  a  note  of  $215,  which  had  been  on  interest  two  years, 
there  would  then  be  due  $227.90 ;  that  is,  $215  principal-f 


164 


SIMPLE  INTEREST. 


$12.90  interest^ $227.90,  the  amount.    Therefore,  the  amount 
is  found  by  adding  the  principal  and  interest  together. 

2.  What  is  the  interest  of  $47.86  for  3  years,  at  6  per  cent, 
per  annum  ? 

OPERATION. 

47.86 
.06 


2.87   1  6 = interest  for  one  year. 
3 


$8.61   4  8= interest  for  three  years. 

To  understand  why  4  figures  are  cut  off  in  this  sum,  the 
scholar  should  remember,  that  1  cent  is  TJ^  of  a  dollar ;  and 
consequently,  6  cents  are  y§^  of  a  dollar,  which  is  "the  same  as 
.06.  (See  introductory  remarks  to  Decimal  Fractions.) 

3.  What  is  the  interest  of  $72.72  for  4  years,  at  6  per  cent.  ? 
Ans.  $17.45.+ 

4.  What  is  the    amount   of  $456  for   two  years?     Ans. 
$510.72. 

5.  What  is  the  interest  of  $146.31   for  five  years,  at  6  per 
cent.  1     Ans.  43.893. 

6.  What  is  the  interest  of  $24.91  for 6  years,  at  5  per  cent,  ? 
Ans.  $7.473. 

7.  What  is  the  interest  of  $222.46  for  three  years,  at  3  per 
cent.  ?     Ans.  $26.695.+ 

8.  What  is  the  amount  of  $42  for  six  years,  at  3  per  cent.  ? 
Ans.  $49.56. 

9.  What  is  the  amount  of  $566.33  for  one  year,  at  5  per 
cent.?     Ans.  $594.646.+ 

10.  What  is  the  amount  of  $1567  for  9  years,  at  2  per  cent.  ? 
Ans.  $1849.06. 

CASE  2d. — WHEN  THE  TIME  CONSISTS  OF  YEARS  AND  MONTHS. 

Lawful  interest  in  the  New  England  states  is  6  per  cent, 
per  annum ;  that  is,  it  is  6  cents  for  12  months,  or  J  cent  per 
month  on  a  single  dollar.  Therefore, 

RULE. — Reduce  the  given  years  to  months,  and  add  the  given 
months ;  then  with  half  this  number  of  months  as  a  multiplier  t 
multiply  the  given  sum.  The  product  will  be  the  interest  for  the 
whole  time. 


SIMPLE  INTEREST.  165 

Note  2d. — It  is  obvious  that  half  the  whole  number  of  months 
in  the  given  time,  is  the  same  as  the  number  of  cents  on  a 
dollar  for  the  whole  time,  when  the  interest  is  at  6  per  cent. ; 
for,  from  the  above  remark,  the  interest  of  one  dollar  for  one 
month  is  evidently  £  per  cent.,  therefore,  the  whole  number  of 
months  divided  by  2,  must  determine  the  number  of  cents  on 
each  dollar  for  the  whole  time.  If,  in  taking  half  the  number 
of  months,  there  be  an  odd  one,  the  interest  for  that  odd 
month  will  be  £  cent,  or  5  mills. 

Ex.  1.  What  is  the  interest  of  $220  for  2  years  and  6 
months,  at  6  per  cent.  ? 

2  yr.  6  months =30  months,  and  30-^-2  =  15.  The  interest 
on  each  dollar  for  the  whole  time  is,  therefore,  15  cents.  Con- 
sequently, $33  is  the  interest  of  the  given  sum. 


22 


$33.00  Ans. 

2.  What  is  the  interest  of  $756.20  for  I  year  and  3  months, 
at  6  per  cent.  ? 

lyr.  3  mo.  =  15  months  ;  therefore,  7£  cents,  or  7  cents,  5 
mills,  is  the  interest  on  each  dollar  for  the  whole  time.  There- 
fore, 

*    75  6.2  0 
.075 


3781 00 
529340 

5  6.7  1  5  00     Ans.  $56.715. 

Note  3d. — If  the  interest  be  required  at  some  other  than  6 
per  cent.,  first  cast  it  at  6  per  cent,  by  the  preceding  rule.  Then 
divide  the  interest  so  found  by  6,  and  the  quotient  will  be  the 
interest  of  the  given  sum  at  one  per  cent,  for  the  time  speci- 
fied ;  and  this  multiplied  by  the  given  per  cent.,  will  be  the  re- 
quired interest. 


166  SIMPLE  INTEREST. 

3.  What  is  the  interest  of  $656  at  8  per  cent,  per  annum, 
for  3  years  and  4  months  ? 

3yr.  and  4  mo.=:40  months,  and  40-i-2z=20,  the  per  cent, 
for  the  whole  time.     Therefore, 
656 
.2  0 


6)131.2  0= interest  at  6  per  cent. 

2  1.866  6= interest  at  1  per  cent. 
8 


174.932  S^interest  at  8  per  cent.,  viz. 
$174.932.+ 

4.  What  is  the  interest  of  $37.50  for  3  years  and  6  months, 
at  6  per  cent.  ?     Ans.  $7.875. 

5.  What  is  the  interest  of  $672  for  3  years  and  8  months,  at 
6  per  cent.  ?     Ans.  147.84. 

6.  What  is  the  interest  of  $372  for  2  years  and  11  months, 
at  5  per  cent.  ?     Ans.  $54.25. 

7.  What  is  the  interest  of  $215.34  for  4  years  and  6  months, 
at  3£  per  cent.  ?     Ans.  33.916.+ 

8.  What  is  the  interest  of  $350  for  2  years  and  6  months, 
at  6  per  cent.  ?    Ans.  $52.50. 

CASE  3d. — WHEN  THE  GIVEN  TIME  CONSISTS  OF  YEARS, 

MONTHS,  AND  DAYS. 

It  was  shown  in  the  preceding  case,  that  the  interest  of  $1 
for  one  month  was  £  cent =5  mills,  whenever  the  rate  is  6 
per  cent.  Now  since  30  days  are  always  allowed  for  a  month 
in  computing  interest,  5  mills  also  equals  the  interest  for! 30 
days;  and  30^-5=6,  the  number  of  days  required  for  one  dol- 
lar at  6  per  cent,  per  annum,  to  gain  one  mill.  If,  therefore, 
the  number  of  days  given,  be  divided  by  6,  the  quotient  will 
be  the  number  of  mills  to  which  the  interest  of  one  dollar 
will  amount  during  that  time.  Therefore, 

RULE. — Find  the  interest  of  one  dollar  for  the  given  time,  by 
allowing  half  a  cent  for  every  month,  and  one  mill  for  every  six 
days ;  the  sum  thus  obtained  will  be  the  per  cent,  for  the  whole 
time.  Multiply  the  principal  by  this,  and  the  product  will  be 
the  interest. 


SIMPLE  INTEREST.  167 

Note  4th. — If  the  given  days  be  less  than  6,  or  if  they  be 
more,  and  in  taking  a  sixth  part  of  them,  a  number  remain, 
(which  in  all  cases  will  be  less  than  6,)  the  interest  for  those 
days  will  be  a  fraction  of  a  mill ;  that  is,  it  will  be  as  many 
sixths  of  a  mill  as  there  are  days. 

Note  5th. — This  rule  always  gives  the  interest  at  6  per 
cent,  per  annum.  If,  therefore,  any  other  per  cent,  be  required, 
proceed  as  directed  in  Note  3d. 

Ex.  1.  What  is  the  interest  of  $150  for  I  year,  3  months, 
and  15  days,  at  6  per  cent,  per  annum  ? 

Solution  :  1  year  and  3  months=15  months;  the  interest, 
therefore,  for  that  time  is  15  half  cents =7%  cents,  or  7  cents, 
5  mills;  and  the  interest  for  15  days  is  1 5  -h6 =2%  mills ; 
therefore,  7  cents,  5  mills 4-2^-  mills=:7  cents,  7£  mills,  which 
is  the  interest  of  one  dollar  at  6  per  cent,  for  the  whole  time. 
Consequently, 

1  50 
.0  7  7i 


1050 
1050 

7  5 

$  1  1.6  2  5= Ans.  or  interest  required. 

2.  What  is  the  interest  of  $320  for  3  years,  4  months,  and 
18  days,  at  6  per  cent,  per  annum  ? 

Solution  :  3  yr.  4  mo. =40  months.  The  interest  for  this 
time  is  .20  per  cent,  and  for  18  days  it  is  18-f-6  =  3  mills  ; 
therefore,  20  cents  and  3  mills  is  the  interest  of  one  dollar  for 
the  whole  tinrc.  Hence, 

320 
.203 

960 
6400 


$64.96  O^interest  required,  viz.  $64.96. 

3.  What  is  the  interest  of  $75  for  1  year,  6  months,  and  12 
days,  at  6  per  cent,  per  annum  ?     Ans.  $6.90. 

4.  What  is  the  interest  of  $162  for  1  year,  8  months,  and  15 
days,  at  6  per  cent.  ?     Ans.  $16.605. 


168  SIMPLE  INTEREST. 

5.  What  is  the  interest  of  $350  for  2  years,  3  months,  and 
5  days,  at  6  per  cent.  ?     Ans.  $47.54.  + 

6.  What  is  the  interest  of  $275  for  3  years,   6  months, 
and  12  days,  at  4  per  cent.  ?     Ans.  $38.866.+ 

7.  What  is  the  interest  of  $560  for  9  months  and  20  days, 
at  6  per  cent.  ?     Ans.  $27.066.  + 

8.  What  is  the  interest  of  $420  for  2  years,  1  month,  and 
27  days,  at  6  per  cent.  ?     Ans.  $54.39. 

Note  6th. — If  the  interest  of  $100  for  one  year  be  $6,  the* 
interest  of  $200  for  the  same  time,  is  obviously  $12.  Again, 
if  the  interest  of  $100  for  one  year  be  $6,  for  2  years  it  must 
be  twice  as  much,  viz.  $12  ;  and  for  3  years,  three  times  as 
much,  viz.  $18.  In  casting  interest,  we  are  therefore  to  regard 
not  only  the  principal,  but  also  the  time  during  which  it  is  on 
interest.  This  may  help  the  scholar  to  understand  the  follow- 
ing rule  for  casting  interest  by  canceling. 

RULE  FOR  CANCELING AND   FIRST,  WHEN  YEARS  ONLY  ARE 

GIVEN. — Draw  a  horizontal  line,  and  write  the  sum  on  which  the 
interest  is  to  be  cast,  above  it.  On  the  right  of  this,  also  above 
the  line,  place  the  given  time  in  years,  and  the  rate  per  cent. 
Then  place  $100  below  the  line.  Proceed  to  cancel,  <$fC. ;]  the 
sum  obtained  will  be  the  required  interest . 

Ex.  1.  What  is  the  interest  of  $300  for  3  years  at  6  per 
cent.  ? 

300.  3.  6 
Statement,          — . 

It  is  obvious,  that  the  interest  of  $300  for  one  year,  will  be 
three  times  as  much  as  the  interest  of  $100  for  the  same  time ; 
and  that  for  3  years,  it  will  be  three  times  as  great  as  for  1 

3 

»•««     o     e 

year.     The  above  statement  canceled :  — ^~~  '•>  an^  3  x  3  x 

6=$54,  Ans. 

2.  What  is  the  interest  of  $350  for  4  years,  at  5  per  cent, 
per  annum  ? 

Statement,   ^'^  5. 

2 

Canceled,  — ~,  and  35x2  =  $70,  Ans. 
S 


SIMPLE  INTEREST.  169 

3.  What  is  the  interest  of  $500  for  5  years,  at  9  per  cent, 
per  annum.     Ans.  $225. 

4.  What  is  the  interest  of  $240  for  6  years,  at  4  per  cent, 
per  annum  ?     Ans.  $57.60. 

AGAIN,   WHEN  THE    GIVEN   TIME    CONSISTS    OF    YEARS    AND 

MONTHS. 

RULE. — Having  reduced  the  given  time  to  months,  arrange 
the  terms  as  in  the  preceding  rule,  and  in  addition  to  the  terms 
there  introduced,  place  12,  the  number  of  months  in  one  year, 
below  the  line ;  cancel,  fyc. 

Note  7th. — If  the  interest  of  $100  for  one  year  be  $6, 
for  two  years  it  would  be  $12,  &c.  (See  preceding  note.)  Con- 
sequently, for  one  year  and  six  months,  it  would  be  J  of  $6=^-|, 
and  for  two  years  and  nine  months,  V^yf  of  $6,  &c.  These 
fractional  expressions  are  obtained  by  reducing  the  given  time 
to  months,  for  the  numerator,  and  making  12,  the  number  of 
months  in  a  year,  the  denominator.  Hence  we  see  the  reason 
of  the  above  rule. 

Ex.  5.  What  is  the  interest  of  $360  for  2  years  and  6 
months,  at  6  per  cent,  per  annum?  Solution :  2  years,  6  months 

O/*A       OA       £ 

—  30  months;  therefore,-      r^— .     It  will  be  observed  that 

100.   la.  ,-. 

the  whole  time  is  ^  of  a  year. 

3 
Canceled,  ^  ^    6,  and  3  x  3  x  6  =  $54,  Ans. 

IWU,.    tJS 

6.  What  is  the  interest  of  $440  for  2  years  and  4  months, 
at  6  per  cent,  per  annum  ? 

The  time  =rf  |  of  a  year,  therefore,  ^?|^.     Ans.  861.60. 

7.  What  is  the  interest  of  $150  for  1  year  and  6  months,  at 
6  per  cent,  per  annum  ? 

150.  18.  6  ...  _A 

Statement,    ,     1c5 — .     Ans.  $13.50. 

JLUU.    1/e 

8.  What  is  the  interest  of  $25.32,  for  9  months,  at  4  per 
cent,  per  annum  ? 

or;  QO    q    A 

Statement,  *£££*,  An*  $0.759.+ 

J.UO.  J« 

15 


170  SIMPLE  INTEREST. 

9.  What  is  the  interest  of  $375.75  for  4  years  and  9  months, 
at  7  per  cent.  ?  Ans.  $124.936.+ 

10.  What  is  the  interest  for  $1500  for  3  years  and  4  months, 
at  6  per  cent.  ?     Ans.  $300. 

11.  What  is  the  interest  of  $175  for  4  years  and  2  months, 
at  8 per  cent.?     Ans.  $58.333.+ 

LASTLY,  WHEN  THE   TIME  CONSISTS  OF  YEARS,  MONTHS,  AND 

DAYS. 

RULE. — Reduce  the  years  and  months  to  months,  and  the 
days  to  the  fraction  of  a  month ;  then  placing  this  fraction  on 
the  right  of  the  months,  reduce  the  whole  to  an  improper  fraction. 
This  expression  will  represent  the  whole  time  in  months ;  there- 
fore, having  arranged  the  other  terms  as  before,  write  its  nume- 
rator above  the  horizontal  line,  and  its  denominator  below,  and 
proceed  as  before  directed. 

Note  8th. — As  the  fractional  expression  of  the  whole  time 
obtained  by  the  preceding  rule  represents  that  time  in  months, 
12,  the  number  of  months  in  one  year,  must  be  placed  below 
the  line,  as  before  directed,  to  reduce  those  months  to  years. 

Ex.  1.  What  is  the  interest  of  $150  for  1  year,  3  months, 
and  1 5  days,  at  6  per  cent.  ?  1  year  and  3  months  —  1 5  months ; 
and  15  days^i  of  a  month;  therefore,  15£  months  is  the 
whole  time  given,  and  this  reduced  to  an  improper  fraction, 
equals  ^  of  a  month.  Agreeably  to  the  rule,  we  then  have 

the  following 

.      150.  31.   6 
Statement,  viz.  1QQ    g   lg. 

3 
Canceled,  ^~r|;    then,  31  x 3  =  93,    and  2x2x2  =  8' 

isviW,.    <*.     i*a- 

2  2 

Therefore,  93  -H-8  =  $11. 625,  Ans. 

2.  What  is  the  interest  of  $320  for  3  years,  4  months,  and 
18  days,  at  6  per  cent,  per  annum?  3  years  and  4  months  = 
40  months,  and  18  days  =  i-|=f  of  a  month.  The  whole  time 
is  therefore  40|  =-§ -  of  one  month. 

320.  203.  6    . 
Therefore,  — — — —  is  the  statement. 


SIMPLE  INTEREST.  171 

S68 

'Q^A      9f^^      I? 

The  same  canceled,  &c.  is  ~—  ^  p^  ;  203x8  =  1624;  and 

5  '  S 

5x5  =  25.     Hence,  1624-^25  =  $64.96,  Ans. 

3.  What  is  the  interest  of  875  for  1  year,  6  months,  and  1  0 
days,  at  6  per  cent.  1     Ans.  $6.875. 

'  55' 


The  time^V  of  a  month,  therefore,  —'  Q'  .    . 

1UU.    o.     1-w 

Note  9th.  —  Ininterest,30daysare  always  allowed  for  a  month. 

In  solving  the  following  sums,  the  scholar  can  apply  either 
of  the  preceding  rules. 

4.  What  is  the  interest  of  $4.20  for  2  years,  1  month,  and 
27  days,  at  6  per  cent.  1    Ans.   $54.39. 

5.  What  is  the  interest  of  $80  for  1  year,  5  months,  and  12 
days,  at  6  per  cent.  1     Ans.  $6.96. 

6.  What  is  the  amount  of  $275  for  3  years,  6  months,  and 
12  days,  at  4  per  cent.  1     $313.866.+ 

7.  What  is  the  amount  of  $560  for  9  months  and  20  days, 
at  6  per  cent.  1     Ans.  $587.066.+ 

8.  What  is  the  interest  of  $50,11  for  1   year  and  21  days, 
at  6  per  cent.  1     Ans.  $3.18.+ 

9.  What  is  the  interest  of  $340.50  for  3  months  and  1  day, 
at  6  per  cent.  1     Ans.  $5.16.+ 

10.  What  is  the  interest  of  $90  for  1  year,  2  months,  and  6 
days,  at  6  per  cent.  1     Ans.  $6.39. 

11.  What  is  the  interest  of  $4119.20  for  1  year  and  5  days, 
at  6  per  cent.  1     Ans.  $250.584.+ 

12.  What  is  the  interest  of  $23.08  for  3  years,   6  months, 
and  22  days,  at  6  per  cent.  1     Ans.  $4.93.+ 

13.  What  is  the  interest  of  $439.50  for   1  year,   11  months, 
and  9  days,  at  6  per  cent.  ?     Ans.  $51.20. 

14.  What  is  the  amount  of  $28  for  9  years,  8  months,   and 
•3  days,  at  6_per  cent.  1     Ans.  $44.254. 

15.  What  is  the  amount  of  $42  for  5  years,  5  months,   and  9 
days,  at  5  per  cent.  ?     Ans.  $53.427. 

16.  What  is  the  amount  of  $50.50  for  1  year  and  3  months, 
at  3  per  cent.  1     Ans.  $52.393.  + 

17.  What  is  the  amount  of  $300  for  16  years   and  8  months, 
at  6  per  cent.  ?     Ans.  $600. 

18.  What  is  the  interest  of  $375.75  for  4  years  and  9  months, 
at  7  per  cent.  ?     Ans.  124.936.  + 


172  SIMPLE  INTEREST. 

19.  What  is  the  interest  of  $3.75  for  12  years,  at  6  per  cent, 
per  annum  ?     Ans.  $2.70. 

20.  What  is  the  amount  of  $75  for  6  years  and  3  months,  at 
6  per  cent.  ?     Ans.  $103.125. 

21.  What  is  the  interest  of  $63  for  1  year  and  3  months,  at 

6  per  cent.  ?     Ans.  $4.725. 

22.  What  is  the  interest  of   $156  for  2  years  and  4  months, 
at  8  per  cent.  ?     Ans.  $29.12. 

23.  What  is  the  amount  of  $650  for  3  years  and  2  months, 
at  6  per  cent.  ?     Ans.  $773.50. 

24.  What  is  the  amount  of  $128.30  for  1  year  and  9  months, 
at  9  per  cent.  ?     Ans.  $148.507. 

25.  What  is  the  interest  of  $33.50  for  2  years  and  6  months, 
at  5  per  cent.  ?     Ans.  $4.1875. 

26.  What  is  the  interest  of  $150  for  2  years,  7  months,  and 

7  days,  at  6  per  cent.  ?     Ans.  $23.425. 

27.  WThat  is  the  interest  of  $730  for  5  years,  9  months,   and 
12  days,  at  6  per  cent.  ?     Ans.  $253.31. 

28.  What  is  the  interest  of  $875.49  for  5  years,   8  months, 
and  20  days,  at  6  per  cent.  ?     Ans.  $300.58.  -f 

29.  What  is  the  amount  of  $630  for  8  months,  at  6  per  cent.? 
Ans.  $655.20.      . 

30.  What  is  the  amount  of  $734%>for  1  year  and  4  months, 
at  6  per  cent.  ?     Ans.  $7929.36. 

31.  What  is  the  amount  of  $750  for  9  months,  at  7  per  cent.  ? 
Ans.  $789.375. 

32.  What  is  the  amount  of  $375  for  5  months  and  15  days, 
at  6  per  cent.  ?     Ans.  $385.31. 

33.  What  is  the  amount  of  $460.50  for  4  months,  at  6  per 
cent.?     Ans.  $469.71. 

34.  What  is  the  amount  of  $230.25,  for  8  months,  at  7  per 
cent.?     Ans.  $240.995. 

35.  What  is  the   amount  of  $764.50  for  3  years   and   10 
months,  at  6  per  cent.  ?     Ans.  $940.335. 

CASE  4th. — To  FIND  THE  INTEREST  AT  six  PER  CENT.  FOR  ANY 

NUMBER  OF  DAYS. 

In  computing  interest,  the  month  is  reckoned  at  30  days, 
and  the  interest  of  one  dollar  for  that  time  at  6  per  cent.,  as 
has  already  been  shown,  is  £  cent  or  5  mills  ;  hence  for  twice 
that  time,  or  60  days,  it  would  be  just  one  cent  for  every  dol- 
lar on  interest.  We  therefore  have  the  following  rule  ; 


SIMPLE  INTEREST.  173 

RULE. — Consider  the  number  of  dollars  given  so  many  cents, 
and  reduce  these  cents  to  dollars  again  by  removing  the  point  of 
separation  two  places  to  the  left ;  the  result  will  be  the  interest  of 
the  given  sum  for  60  days.  Then,  if  the  given  days  be  more 
or  less  than  60,  add  to,  or  subtract  from,  the  interest  of  GO  days, 
such  parts  of  itself  as  the  given  days  require. 

Ex.  1.  What  is  the  interest  of  $450.82  for  93  days,  at  6 
per  cent,  per  annum  1 

SOLUTION. 

4.5082   =  interest  for  60  days.  (See  the  rule.) 
2.2541    ^interest  for  30  days,  being  half  the  int.  for  60  days. 
22541  ^interest  for  3  days,  being  ^  of  the  int.  for  30  days. 

$6.98771  ^interest  for  93  days. 

2.  What  is  the  interest  of  $4562  for  45  days,  at  6  per  cent, 
per  annum  ? 

SOLUTION. 
45.62   ^interest  for  60  days. 

22.81    ^interest  of  30  days,  or  one  half  the  int.  of  60  days. 
11. 405= interest  of  15  days,  or  one  half  the  int.  of  30  days, 

$34.215=interest  of  30  +  15  =  45  days. 

In  computing  by  this  rule,  12  months  of  only  30  days  each, 
are  allowed  for  the  year,  equal  to  360  days.  It  consequently 
gives  Jg-  part  more  than  6  per  cent,  interest.  On  small  sums, 
and  for  short  intervals,  however,  the  difference  is  trifling. 

3.  What  is  the  interest  of  $420.72  for   120  days,  at  6  per 
cent,  per  annum  ?     Ans.  $8.414.+ 

4.  What  is  the  interest  of  $56.74  for  25  days,  at  6  per  cent, 
per  annum  ?     Ans.  $0.236.+ 

5.  What  is  the  interest  of  $156.36  for  96  days,  at  6  per 
cent,  per  annum?     Ans.  $2.50.+ 

6.  What  is  the  interest  of  $1000  for  29  days,  at  6  per  cent, 
per  annum  ?     Ans.  $4.833.+ 

7.  What  is  the  interest  of  $204  for  40  days,  at  6  per  cent, 
per  annum  1     Ans.  $1.36. 

8.  What  is  the  interest  of  $472  for  18  days,  at  6  per  cent 
per  annum?     Ans.  $1.416. 

15* 


174  SIMPLE  INTEREST.  . 

Banking. — When  a  promissory  note  is  presented  at  a  bank- 
ing institution,  if  properly  endorsed  or  otherwise  secured,  it  is 
received  by  the  officers  of  the  bank-  as  security,  and  so  much 
money  in  their  own  notes  is  given  in  return  as  is  equal  to  the 
face  of  the  note  after  the  interest  is  deducted  for  3  days  more 
than  the  whole  time  till  payment  is  promised.  Hence,  if  the 
time  specified  be  60  days,  the  interest  on  the  face  of  the  note 
for  63  days  is  deducted,  and  the  balance  drawn  from  the  bank  ; 
then  at  the  expiration  of  the  63  days,  the  whole  face  of  the 
note  is  due.  The  3  days  added  to  the  specified  time  of  pay- 
ment, are  called  "  days  of  grace ." 

The  preceding  rule  is,  therefore,  a  convenient  one  for  bank- 
ing institutions. 

In  solving  the  following  sums,  the  scholar  will  allow  "  3 
days  of  grace"  that  is,  he  will  find  bank  discount  for  3  days 
more  than  are  specified  in  the  sum. 

Ex.  1.  What  is  the  bank  discount  on  $256  for  30  days,  and 
grace  ? 

SOLUTION. 
2  . 5  6= discount  for  60  days. 


1  .  2  8= discount  for  30  days. 
.  1  2  8= discount  for  3  days  grace. 

$1.40  8=required  discount. 

Therefore,  $256 — $1.408  =  $254. 592,  the  sum  to  be  drawn 
from  the  bank. 

2.  How  much  money  would  be  drawn  from  the  bank  on  a 
note  of  $650,  payable  in  90  days  ? 

6  .  5  0  =  discount  for  sixty  days. 
3  .2  5 rediscount  for  thirty  days. 
.  3  2  5= discount  for  three  days. 


10.07  5= discount  for  ninety -three  days. 
Therefore,    $650 — $10. 075  =  $639.925,   the   money   to    be 
drawn. 

3.  What  is  the  bank  discount  on   $1056.29  for  30  days, 
and  grace?     Ans.  $5.81. 

4.  What  is  the  bank  discount  on   $756  for  90  days,  and 
grace?     Ans.  $11.718. 


LE  INTEREST.  175 

5.  What  is  the  barflff&iscount  on  $676.19  for  25  days,  and 
grace?     Ans.  $3.155.+ 

6.  How  much  money  «iay  be  drawn  on  a  note  of  $692, 
payable  70  days  after  date,  if  discounted  at  a  bank ?     Ans'. 
$683.581. 

7.  How  much  money  may  be  drawn  on  a  note  of  $1567.89, 
payable  120  days  from  date?     Ans.  $1535.748. 

8.  What  is  the  bank  discount  on  $542.78  for  90  days,  and 
grace?     Ans.$8Al3.+ 

9.  What  is  the  bank  discount  on  $195.77  for  60  days,  and 
grace?     Ans.  $2.055.+ 

10.  How  much  money  may  be  drawn  on  a  note  of  $726, 
payable  80  days  from  date  ?     Ans.  $715.957. 

CASE  5th. — WHEN  THE  MONEY  ON  WHICH   THE  INTEREST  is 

TO  BE  CAST,  IS  IN  POUNDS,  SHILLINGS,  AND  PENCE. 

RULE. — Reduce  the  shillings  and  pence  to  the,dfaimal  of  a 
pound,  (see  Case  3d,  Decimal  Fractions,)  and  castmfte  interest 
in  the  same  manner  as  when  the  principal  consists  of  dollars, 
cents,  and  mills.  The  decimal  part  of  the  answer  may  then  be 
reduced  to  shillings  and  pence  by  Case  4th,  Decimal  Fractions. 

Ex.  1.  What  is  the  interest  on  42  £.  16s.  6  d.  for  1  year 
and  6  months,  at  6  per  cent,  per  annum  ? 
SOLUTION. 
12 

20  § 

.8  2  5= the  decimal  value  of  16s.  6d. 

1  yr.  6mo.r=;18  months,  and  18-^2  =  9,  the  per  cent,  for  the 
whole  time.     Therefore, 
4  2.8  2  5  £. 
.0  9 


3.8  5  425  £.=  interest  in  pounds  and  decimals. 
To  find  the  value  of  the  decimal : 
.85425 

2  0 


17.08500 
1  2 

1.02000 
The  answer,  therefore,  is  3  j£.  17s.  Id. 


176  SIMPLE  INTERES 


\afi 


2.  What  is  the  interest  of  55  £.  lJP6d.  for  2  years,  6 
months,  at  6  per  cent.  ? 

55  £.  15  s.  6d.  ^55.775  £.    And2  yr.  6  mo.  =  30  months  = 
15  per  cent,  for  the  whole  time.     Therefore, 
55.775 
.1  5 


278875 
55775 

8.3  6625. 

To  find  the  value  of  the  decimal : 
.36625 
2  0 


7.3  2  5  0  0=  shillings. 
12 


3.9000  Oz=pence. 
4 


3.6  0  0  0  0=farthings. 
Therefore,  8  £.  7  s.  3  d.  3  qr.  Ans. 

3.  What  is  the  interest  of  21  £.  18  s.  4  d.,  for  3  years  and 
4  months,  at  6  per  cent.  ?     Ans.  4  £.  7s.  8  d. 

4.  What  is  the  amount  of  156£.  9s.  3  d.  for  1  year  and  9 
months,  at  6  per  cent.  1     Ans.  172  £.  17s.  9  d.  3  qr.+ 

5.  What  is  the  amount  of  27  £.  2s.  6d.   3  qr.  for   1  year 
and  10  months,  at  6  per  cent.  ?     Ans.  30  £.  2  s.  2  d.  3  qr.+ 

6.  What  is  the   interest   of  36  £.  15s.  for  2  years  and  3 
months,  at  6  per  cent.  ?     Ans.  4£.  19  s.  2  d.  2.8  qr. 

7.  What  is  the  interest  of  45  £.  10  s.  for  8  months,  at  6  per 
cent.  ?     Ans.  I  £.  16  s.  4  d.  3.2  qr. 

8.  What   is   the  interest  of  9  £.  12  s.  6  d.  for  2  years,  4 
months,  and  12  days,  at  6  per  cent.  ?  Ans.  \  £>.  7s.  4  d.  -f- 

CASE  6th. — WHEN  INTEREST  is  REQUIRED  ON  NOTES  OR  BONDS 

ON  WHICH  PARTIAL  PAYMENTS  HAVE  BEEN  MADE. 

RULE. — Cast  the  interest  on  the  principal  at  the  given 
rate  per  cent,  up  to  the  time  of  the  first  payment ;  then,  if  the 
payment  exceed  the  interest,  deduct  the  excess  from  theprincipal ; 


SIMPLE  INTEREST.  177 

but  if  it  be  less,  set  both  payment  and  interest  aside,  and  cast 
the  interest  on  the  same  principal  to  the  time  of  the  next  pay- 
ment, or  to  the  time  of  some  payment,  which,  when  added  to  the 
preceding  payments,  will  exceed  the  sum  of  interest  then  due, 
and  deduct  the  sum  of  these  payments  from  the  amount  of 
the  principal.  The  remainder  will  form  a  new  principal,  with 
which  proceed  as  before,  till  the  time  of  settlement. 

1.  For  value  received,  I  promise  to  pay  A.  B.  &  Co.,  or 
order,  fifteen  hundred  dollars  on  demand,  with  interest. 
Jan.  1,  1825.  JOHN  JAMES. 

On  this  note  are  the  following  endorsements  :  Oct.  1,  1825, 
three  hundred  dollars.  July  1,  1827,  four  hundred  and  fifty 
dollars.  Sept.  1,  1828,  six  hundred  and  fifty  dollars.  What 
was  due  on  settlement,  July  1,  1830? 

The  principal  on  interest  from  Jan.-  1,  1825, $1500.00 

Interest  from  Jan.  I,  1825,  to  Oct.  1,  of  the  same  year,  9  mo.        67.50 

1567.50 
The  payment  being  more  than  the  interest,  deduct  it,        .     .      300.00 

Remainder,  forming  a  new  principal, 1267.50 

The  interest  from  Oct.  1, 1825,  to  July  1,  1827,  1  yr.  and  9  mo.  133.87 

1400.587 
Deduct  second  payment,  because  it  is  more  than  the  interest,    450.000 

Remainder,  forming  a  new  principal, 950.587 

The  interest  from  July  1,  1827,  to  Sept.  1, 1828,  1  yr.  and  2  mo.  66.541 

1017.128 
Deduct  the  third  payment, 650.000 


Remainder,  forming  a  new  principal, ;    .    367.128 

The  interest  from  Sept.  1,  1828,  to  July  1,  1830,  the  time  of 

settlement, 40.384 

Ans.  $407.512 

SOLUTION  BY  CANCELING. 

yr.  m.  d. 
J825  10  1 
1825  1  1 

9     0  time  till  first  payment. 


178  SIMPLE  INTEREST. 

aftd  15x9-^-2  =  $67.50,  the  interest  till  first  payment ;  and 
1900  + 67.50  =  $1567.50,  amount;  and  1567.50—300  = 
$1267.50,  the  new  principal. 

yr.  mo.  d. 
1827  7  1 
1825  10  1 


1       9     0=time  from  1st  to  2d  payment. 

1267.50.   21.   6    "  1267.50.    21.    & 

Statement,    -      --    -.     Canceled,   _ 


2 

1267.50x21-^-200  =  133.087,  the  interest  till  2d  payment; 
then,  1267.504-133.087  =  1400.587,  and  1400.587—450  = 
950.587,  the  new  principal. 

yr.  m.  d. 
1828  9  1 
1827  7  1 


1     2     0=time  from  2d  to  3d  payment. 

7 

c.  4  ,.    950.587.  14.  6       ^         ,    ,    950.587.  14.  S 

Statement,  -___,     Canceled,   -^  -^-  ; 


and  956.587x7-^-100=66.541,  interest  till  the  3d  payment, 
and  950.587+66.541  =  1017.128,  and  1017.128—  650  =  367.- 
128,  the  new  principal. 
yr.       m.    d. 

1830     7     1 

1828     9     1 


110     0=time  from  3d  payment  to  settlement. 

11 

367.128.  22.  6     ^         ,    ,    367.128.  SS.  & 
Statement,  —     —  -—  .    Canceled,  --    --    —  ; 


and  376.128x11-^-100=40.384,  interest  till  time  of  settle- 
ment, and  367.  128  +40.  384  =  $407.5  12,  answer,  or  sum  due 
on  settlement. 

2.  I  have  in  my  possession  a  note,  dated  April  15,  1833,  for 
$2150.25,  on  which  are*  the  following  endorsements  :  Nov.  8, 
1834,  $500.00;  Sept.  1,  1835,-  $723.64;  January  1,  1837, 
$378.295  ;  and  Oct.  29,  1837,  $850.00.  What  amount  was 
due  on  this  note,  April  15,  1838?  Ans.  $138.337. 


COMPOUND  INTEREST.  179 

3.  On  a  note  of  $767.95,  given  Dec.  25,  1827,  and  drawing 
interest  after  ninety  days,  were  the  following  endorsements  : 
Jan.  1,  1830,  $75.00  ;  March  25,  1831,  $565.25.     What  was 
due  Jan.  1,1833?     Ans.  $294.118. 

BOSTON,  Jan.  13,  1809. 

4.  On  demand,  I  promise  to  pay  J.  Anderson,  or  order,  one 
thousand  five  hundred  eighty-five  and  331, 0.0  dollars,  with  in- 
terest, for  value  received. 

Received  May  5,  1812,  $863.12. 
Received  May  7,  1814,  $221. 
Received  July  21,  1815,  $1009.03. 

Will  the  scholar  determine  whether  the  note  is  fully  paid  ? 

5.  What  was  due  on  a  note  of  $2100,  dated  June  15,  1820, 
on  settlement,  June  15,  1830,  the  following  sums  beino-  en- 
dorsed on  the  back  of  it,   viz.  June  30,  1824,  $750,  and  Sept. 
30,  1828,  $1200,  on  interest  at  6  per  cent.  ?     Ans.  $1249.527. 

6.  For  value  received  of  A.  B.,  I   promise  to  pay  him,  or 
order,  seven  hundred  and  fifty  dollars,  with  interest  at  6  per 
cent.? 

Jan.  1,1824.  M.   S. 

On  the  above  were  the  following  payments  endorsed :  April 
1,  1826,  one  hundred  and  fifty  dollars;  July  1,  1829,  four 
hundred  and  fifty  dollars.  What  was  due  on  settlement,  Sept. 
1,  1832?  Ans.  $461.71.+ 


COMPOUND    INTEREST. 

Compound  Interest  is  that  which  is  computed  annually, 
and  immediately  added  to  the  principal.  The  amount  of  each 
year,  is  made  the  principal  for  the  succeeding  year. 

RULE. — Cast  the  interest  at  the  given  rate  per  cent,  for  the 
first  year,  by  multiplying  by  that  rate  per  cent.,  and  make  the 
amount  the  principal  for  the  second  year.  Make  the  amount 
of  the  second  year,  principal  for  the  third ;  and  the  amount  of 
the  third,  the  principal  for  the  fourth,  <$~c.,  through  the  whole 
number  of  years.  From  the  amount  thus  obtained,  subtract  the 
principal ;  the  remainder  will  be  the  Compound  Interest. 


180  COMPOUND  INTEREST. 

Ex.  1.  What  is  the  compound  interest  of  $256  for  3  years, 
at  6  per  cent.  ? 

256 
.0  6 

15.36  interest. 
256.00  principal  added. 


271.36  principal  for  2d  year. 
.0  6 


1  6.28  1  6= interest  of  2d  year. 
271.3  6=principal  of  3d  year  added. 

287.6416  principal  for  3d  year. 
.06 


17. 2  5  8496  interest  of  3d  year, 
287.641  6  principal. 


304.900096  amount  of  3d  year. 
256.0  0=the  original  principal  subtracted. 

48.90  compound  interest  on  $256,  for  3  years. 

2.  What  is  the  compound  interest  of  $450  fox  3  years,  at  6 
percent.?     Ans.  $85.957.  + 

3.  What  is  the  compound  interest  of  $50  for. 3  years,  at  5 
per  cent.  ?     Ans.  $7.881 .  -f- 

4.  What  is  the  compound  interest  of  $400,  at  6  per  cent,  for 
4 years?     Ans.  $104.99.  -f- 

5.  What  will   $675   amount  to  at  compound  interest,  in  3 
years  and  6  months,  6  at  per  cent,  per  annum  ?     Ans.  $828.- 
053.+ 

6.  What  is  the  amount  of  &40.20  at  6  per  cent,   compound 
interest,  for  4  years  ?     Ans.  $50.75.+ 

7.  What  is  the  amount  of  $63  at  6  per  cent,  compound  in- 
terest, for  2  yelrs  ?     Ans.  $70.786.+ 

8.  What  is  the  compound  interest  of  $127.85  for  3  years, 
at  6  per  cent.  ?     Ans.  24.42 l.-f 


INSURANCE.  181 


COMMISSION. 

Commission  is  an  allowance  made  by  merchants  and  others 
to  an  agent  for  buying  and  selling  goods.  This  allowance  is 
usually  a  certain  per  cent,  on  the  amount  of  money  received  for 
the  sales  effected,  or  on  that  expended  in  making  purchases. 
The  only  respect  in  which  it  differs  from  interest,  is,  that  in 
computing  commission  no  regard  is  paid  to  time ;  hence, 

RULE. — Multiply  by  the  rate  per  cent. 

Ex.  1.  An  agent  sold  for  his  employer  goods  to  the  value 
of  $1800,  for  which  he  received  5  per  cent. ;  what  was  the 
amount  of  his  commission?  Ans.  $90. 

2.  What  is  the  amount  of  my  commission  for  selling  goods 
to  the  value  of  $975,  it  being  8  per  cent.  ?     Ans.  $78. 

3.  My  agent  sends  me  word,  that  he  has  purchased  goods 
on  my  account  to  the  value  of  $2768  ;  what  will  his  commis- 
sion amount  to  at  6  per  cent.  ?     A?is.  $166.08. 

4.  The  commission  of  $1250  at  10  per  cent,  is  required  ; 
what  is  it?     Ans.  $125. 

5.  An  agent  sells  750  bales  of  cotton,  at  $52  per  bale,  and 
is  to  receive  2^  per  cent,  commission.    How  much  money  will 
he  receive,  and  how  much  will  he  pay  over  to  his  employer  ? 
Ans.    He  will  receive  $975,  and  pay  over  $38025. 

6.  What  will  my  commission  amount  to  at  3  per  cent,  in 
purchasing  goods  to  the  value  of  $7846.90  ?     Ans.  $235.407. 


INSURANCE. 

Insurance  against  the  loss  of  buildings  and  goods  by  fire, 
and  also  of  ships  and  their  cargoes  by  storm,  is  obtained  by 
paying  a  certain  per  cent,  on  the  estimated  value  of  the  prop- 
erty insured. 

16 


182  INSURANCE. 

The  instrument  which  binds  the  contracting  parties  is  called 
the  policy,  and  the  sum  paid  by  the  party  insured  to  the  insuring 
party,  is  called  the  premium. 

RULE. — Multiply  the  estimated  value  of  the  property  insured, 
by  the  per  cent. 

Ex.  1 .  What  is  the  premium  for  the  insurance  of  buildings 
and  appurtenances,  valued  at  $3758.50,  at  \  per  cent.  1  Aus. 
$18.79.+ 

2.  What  is  the   premium  for  insuring  property,  valued  at 
$3600,  against  loss  by  fire,  at  £  per  cent.  ?     Ans.  $27. 

3.  What  is  the  premium  for  insuring  property  valued  at 
$845,  at  i  per  cent.  1     Ans.  $1.69. 

4.  What  is  the  premium  for   the  insurance  of  a  ship  and 
cargo  valued  at  $20500,  at  |  per  cent.  1     Ans.  $68.333.+ 

5.  What  would  be  the  premium  for  insuring  a  ship  and 
cargo  valued  at  $18000,  at|  per  cent.  ?     Ans.  $67.50. 

6.  Insured  my  house  and  out  buildings,  valued  at  $21560.38, 
at  J  per  cent. ;  what  was  the  amount  of  the  premium  ?     Ans. 
$86.24.+ 

QUESTIONS. — What  is  interest  1  How  is  it  computed  1  Suppose  the 
sum  en  interest  be  more  or  less  than  $100,  or  the  time  more  or  less 
than  one  year,  how  must  the  sum  paid  as  interest  compare  ?  Give  the 
illustration.  What  is  the  principal?  What  is  interest  1  What  is 
legal  interest?  What  is  the  legal  rate  per  cent,  in  New  England? 
What  in  New  York  ?  What  in  Louisiana  ?  What  do  you  under- 
stand by  the  amount  ?  The  rate  per  cent,  is  a  decimal  of  how  many 
places  does  it  consist  when  expressed  by  cents  ?  And  when  expressed  by 
mills?  What  is  Case  1st?  What  is  the  rule  for  it  ?  What  is  note  first  ? 
What  is  Case  2d  ?  What  is  the  rule  ?  Give  the  reason  of  the  rule. 
What  is  note  2d  ?  Noie  3d  ?  What  is  Case  3d  ?  What  is  the  rule  1 
Give  the  explanation  which  precedes  the  rule.  What  is  note  4th  ? 
What  is  note  5th  ?  What  is  note  6th  ?  What  is  the  rule  for  can- 
celing ?  What  is  the  rule  for  canceling,  when  the  time  consists  of  years 
and  months  ?  What  is  Note  7th  ?  What  is  the  rule  for  canceling, 
when  the  time  consists  of  years,  months,  and  days  ?  What  is 
Note  8th  ?  What  is  Note  9th  ?  What  is  Case  4th  ?  What  is  the 
rule  ?  What  explanation  precedes  the  rule?  What  is  said  relative 
to  banking  institutions  ?  What  is  Case  5th  ?  What  is  the  rule  for  it  ? 
What  is  Case  3d  of  Decimal  Fractions?  What  is  Case  4th  of 
Decimal  Fractions?  What,  is  Case  6th?  What  is  the  rule  for  it? 
Will  the  scholar  now  inform  me  why  it  is  correct  to  multiply  by  one 
half  of  the  even  number  of  months  in  the  given  time,  in  casting  inte- 
rest at  6  per  cent.  ?  What  is  Compound  Interest  ?  What  is  the  rule 
for  it?  What  is  Commission  ?  In  what  respect  does  it  differ  from 
simple  interest?  What  is  the  rule?  How  is  insurance  obtained? 
What  is  to  be  understood  by  the  policy?  What  by  the  premium? 
What  is  the  rule  ? 


RATIO.  183 


RATIO. 

Ratio  is  the  relation  which  one  quantity  or  number  has  to 
another  quantity  or  number  of  the  same  kind. 

The  former  of  the  two  numbers,  between  which  the  ratio 
exists,  is  called  the  antecedent,  and  the  latter,  the  consequent. 

The  direct  ratio  of  any  two  numbers  is  obtained  by  dividing 
the  consequent  of  any  couplet  by  the  antecedent,  and  the  in- 
verse ratio,  by  dividing  the  antecedent  by  the  consequent. 

Thus,  the  direct  ratio  of  2  to  4,  is  2,  because  the  antece- 
dent 2,  is  contained  in  the  consequent  4,  two  times  ;  and  the 
inverse  ratio  is  -| ,  because  4,  the  consequent,  is  contained  in 
2,  the  antecedent,  J=r^  of  a  time.  Both  these  expressions 
establish  the  same  general  fact,  viz.  that  one  of  the  given  num- 
bers is  twice  as  large  as  the  other. 

From  the  above,  it  is  obvious  that  ratio  can  exist  only  be- 
tween quantities  of  the  same  kind.  It  would  be  absurd  to  inquire 
how  many  times  3  acres  must  be  repeated  to  equal  12  tons. 

Any  simple  ratio  is  expressed  by  two  dots  placed  between 
the  antecedent  and  the  consequent ;  thus,  2  :  8,  or  5  :  10. 

The  following  four  propositions  require  to  be  carefully 
studied. 

1.  The  ratio  of  any  couplet  is  not   affected  either  by  multi- 
plying or  by  dividing  its  antecedent  and  consequent  by  the  same 
number  ;  for  the  ratio  of  9  :  1 8  is  2  ;  and  the  ratio  of  9TTF  X  2 
=  18  :  36,  is  also  2.     The  same  result  is  obtained  by  dividing 
these  terms  by  any  number  whatever  ;    thus,  9: 18-^3  =  3  :  6, 
and  this  equals  2. 

2.  The  ratio  of  any  couplet  is  multiplied  by  any  number,  by 
multiplying  the  consequent,  or   by  dividing  the  antecedent  by 
that  number.     The  ratio  of  12  to   36,  is  3  ;  if,  however,  the 
consequent  be  multiplied  by  3,  the  ratio  3  will  be  multiplied  by 
the  same  number  ;  thus,  12  :  36  X  3  =  12  :  108  =  9.     The  same 
result  is  obtained  by  dividing  the  antecedent  by  the  same  num- 
ber ;  thus,  12,  the  antecedent,  divided  by  3,  equals  4,  and  the 
ratio  of  4  to  36  is  9,  the  same  as  before. 

3.  The  ratio  of  any   couplet  is  divided  by  any  number,  by 
dividing  the  consequent,  or  by  multiplying  the   antecedent  by 
that  number.     Take  the  ratio  12  :  36,  as  before,  and  let  it  be 
divided  by  2.     If  the  consequent  be  divided,  we  obtain  the 
ratio  12  :  18  =  1^ ;  but  if  the  antecedent  be  multiplied,  we  ob- 
tain the  ratio  24  :  36  =  l£  also. 


184  RATIO. 

4.  If  two  or  more  ratios  be  multiplied  together,  that  is,  if  the 
antecedents  be  multiplied  into  the  antecedents,  and  the  conse- 
quents into  the  consequents,  the  resulting  ratio  is  called  COM- 
POUND RATIO,  and  is  equal  to  the  product  of  the  simple  ratios. 
The  ratio  compounded  of  the  simple  ratios,  2:4,  3:6,  and 
4  :  8,  is  the  ratio,  24  :  192  =  8  ;  but  the  ratio,  2  :  4=2,  also, 
3  :  6=2,  and  4  :  8=2  ;  and  2x2x2  =  8. 

Ex.  1.  What  is  the  ratio  of  6  to  36  ?     Ans.  6. 

2.  What  is  the  inverse  ratio  of  7  to  12  ?     Ans.  f^. 

3.  What  is  the  ratio  of  7  to  42  ? 

4.  What  is  the  inverse  ratio  of  5  to  20  ? 

5.  What  is  the  ratio  of  1  £.  sterling  to  7s.   6  d.  ?     It  is 
obvious  that  the  terms  here  given  must  be  reduced  to  the  same 
denomination,  in  order  to  compare  them,  therefore,  1  £.=240  d. 
and  7  s.  6  d.  =  90  d  .     The  direct  ratio,  therefore,  is  -££$  =  f . 

6.  What  is  the  ratio  of  16  Ib.  to  3  cwt.  1     Ans.  21. 

7.  Multiply  the  terms  3  :  7  by  9,   and  see  how  the   given 
ratio  is  affected. 

8.  Divide  the  terms  9  :  15  by  3,  and  compare  ratios.     (See 
proposition  1st.) 

9.  Multiply  the  ratio  11  :  16  by  5.     Ans.  11  :  80. 

10.  Multiply  the  ratio  9:11  by  3.    Ans.  3:11.  (See  propo- 
sition 2d.) 

11.  Divide  the  ratio  3  : 12  by  6.     Ans.  3  :  2  or  f . 

12.  Divide  the  ratio  3  :  19  by  4.   Ans.  12  : 19.    (See  propo- 
sition 3d.) 

13.  What  is  the  ratio  compounded  of  the  ratios  2:3;  3:4, 
and  4 : 5  ?     Ans.  24  :  60  or  2  :  5. 

Note. — Whenever  the  antecedent  of  any  couplet  is  the 
same  as  the  consequent  of  any  other  couplet,  the  several 
ratios  may  be  reduced  to  one,  by  rejecting  such  antecedents 
and  consequents.  Hence,  the  preceding  simple  ratios  may  be 
reduced  to  a  compound  one,  by  rejecting  their  similar  terms. 
The  ratio  thus  obtained  is  2  :  5. 

14.  What  is  the  ratio  compounded  of  the  simple  ratios  3:5; 
4 :  5  ;  and  2  : 3  ?  (See  proposition  4th.)  Ans.  24  :  75,  or  8  :  25. 

15.  Multiply  the  ratio  5  :  9  by  4.     Ans.  5  :  36. 

16.  Divide  the  same  ratio  by  3.     Ans.  5  :  3. 

17.  Multiply  the  ratio  12  :  21  by  6.     Ans.  2:21. 

18.  Divide  the  same  by  7.     Ans.  84  :  21,  or  12  :  7. 

19.  Reduce  the  ratios  4:7;  3:1;  and  6:2,  to  a  compound 
ratio.     Ans.  36  :  7. 


PROPORTION.  185 


PROPORTION. 

Equality  of  ratios  constitutes  proportion. 

Each  simple  statement  in  proportion  requires  at  least  two 
equal  ratios,  or  four  terms,  the  first  and  second  of  which  are  of 
one  kind,  and  constitute  one  of  the  ratios ;  and  the  third  and 
fouth  are  of  another  kind,  and  constitute  the  other  ratio. 

If  6  men  earn  12  dollars  in  a  given  time,  36  men  will  earn 
72  dollars  in  the  same  time.  In  this  proposition,  the  two  ratios 
required  to  constitute  a  proportion,  are,  the  ratio  of  6  men  to 
36  men,  and  of  12  dollars  to  72  dollars;  and  the  proposition 
can  be  true  only  on  the  condition  that  these  two  ratios  are 
equal;  and  that  is  actually  the  case,  for  36 -4- 6  =  6,  and  72-r- 
12  =  6.  Therefore,  6  is  the  ratio  of  each  couplet. 

Proportion  is  usually  expressed  by  dots,  thus  :  6  :  36  :  :  12  : 
72,  and  thus  read  :  6  is  to  36  as  12  is  to  72,  or  6  men  is  to 
36  men  as  12  dollars  is  to  72  dollars. 

When  any  four  numbers  are  proportionals,  the  first  and  fourth 
are  called  extremes,  and  the  second  and  third,  the  means ;  and 
the  product  of  the  extremes  must  always  equal  the  product  of 
the  means.  This  is  true  with  regard  to  the  statement  made 
above,  for  6x72  =  432,  and  12x36  =  432. 

Since,  therefore,  these  products  are  always  equal,  if  any 
three  terms  are  given,  two  of  which  bear  a  given  ratio  to  each 
other,  a  fourth  term  may  be  found,  to  which  the  third  given 
term  shall  bear  the  same  ratio  ;  for  the  required  term  must  be 
either  one  of  the  extremes  or  one  of  the  means.  If  it  be  one 
of  the  extremes,  the  product  of  the  means  divided  by  the  given 
extreme,  will  give  the  required  extreme ;  and  if  it  be  one  of 
the  means,  the  product  of  the  extremes  divided  by  the  given 
mean,  will  give  the  required  mean. 

Suppose,  in  the  proposition  stated  above,  it  be  required  to 
find  how  many  dollars  36  men  would  earn  in  a  given  time,  if 
6  men  earn  12  dollars  in  the  same  time. 

Let  the  three  given  numbers  stand  as  before,  thus  :  6  men  : 
36  men :  :  12  dollars  :  what  number  of  dollars  ? 

Now,  since  6,  the  left  hand  term,  multiplied  by  the  required 

term,  must  produce  the  same  number  as  36  multiplied  by  12,  it 

follows  that  36  multiplied  by  12,  and  the  product  divided  by  6, 

will  give    the  required]  number.     Thus,  36x12=432,   and 

16* 


186  PROPORTION. 

432-^-6=72,  the  fourth  term  in  the  preceding  proposition. 
Or,  suppose  it  required  to  find  what  number  bears  the  same 
ratio  to  36  that  12  does  to  72.  The  statement  would  be, 
what  number  :  36  :  :  12  :  72.  The  other  extreme  is  here 
required;  hence,  36x12=432,  and  432-f-72=6,  the  re- 
quired extreme.  Again,  let  one  of  the  means  be  required, 
thus  :  6  is  to  what  number  as  12  is  to  72.  The  statement  may 
stand  thus  :  6  :  — :  :  12  :  72.  One  of  the  means  is  here  re- 
quired, and  the  extremes  are  given;  therefore,  72x6  =  432, 
and  432-^6  =  36,  the  required  mean.  Lastly,  let  the  other 
mean  be  required.  We  then  have  the  statement  as  follows  : 
6  : 36  :  :  — :  72.  The  third  term  of  the  statement  is  here  want- 
ing, or  the  other  mean.  Therefore,  72  X  6 =432,  and  432 -^- 
36  =  12,  the  required  mean. 

From  the  preceding  remarks  we  learn  that  operations  in 
simple  proportion  consist  in  having  three  of  the  terms  of  a 
proportion  given  and  a  fourth  term  required.  For  finding 
this  fourth  term,  we  have  the  following  rule  : 

RULE. — Notice  which  of  the  three  terms  given  is  of  the  same 
name  or  kind  as  the  required  term  or  answer,  and  give  it  the 
right  hand  place.  Notice  again  whether  the  term  required  must 
be  greater  or  less  than  this ;  and,  if  it  is  to  be  greater,  place 
the  greater  of  the  two  remaining  terms  next  it,  on  the  left,  for 
the  second  term  of  the  proportion,  and  the  less  number  for  the 
first ;  but  if  it  is  to  be  less,  place  the  less  of  the  two  remaining 
numbers  for  the  second  term,  and  the  greater  for  the  first  term  ; 
then  multiply  the  second  and  third  terms  together  for  a  dividend, 
and  divide  their  product  by  the  first :  the  quotient  will  be  the 
fourth  term,  or  answer,  and  of  the  same  denomination  as  the 
third  term. 

Note  1st. — If  the  third  term  consists  of  different  denominations, 
it  must  be  reduced  to  the  lowest  mentioned  before  stating,  and 
the  fourth  term  will  be  of  the  same  denomination ;  but  if  either 
the  first  or  second  terms  be  of  different  denominations,  they 
must  both  be  reduced  to  the  lowest  mentioned,  before  stating. 

Ex.  1.  If  8  yards  of  cloth  cost  $3.20,  what  will  96  yards 
cost  ? 

Since  8  yards  cost  money,  96  yards  will  cost  money ;  the 
required  term  must,  therefore,  be  money ;  and  must  be  a 
number  to  which  $3.20  will  bear  t}ie  same  ratio  that  8  yards 


PROPORTION.  187 

bears  to  96  yards.     $3.20  is,  therefore,  to  be  made  the  right 
hand  term. 

The  next  inquiry  is,  which  will  cost  most,  8  yards  or  96 
yards  ?  The  answer  to  this  inquiry  places  (in  accordance  with 
the  rule)  the  96  in  the  second  place,  and  the  8  in  the  first,  thus  : 
8  : 96  : :  3.20  :  what  number  dollars  ? 

PERFORMED. 

8  :  9  6  :  :  3.2  0 
96 

1920 
2880 


8)307.20 

_^___^__ 

$  3  8.4  0   Ans. 
Therefore,  8  yards  :  96  yards  :  :  $3.20  :  $38.40. 

2.  If  9  men  earn  72  dollars  in  a  given  time,  how  much  will 
24  men  earn  in  the  same  time  ?     Statement :    9  men :  24  men 
::$72:Ans.  $192.00. 

Sums  of  this  description  may  be  solved  with  peculiar  ease 
by  canceling. 

RULE  FOR  CANCELING. — Notice  which  of  the  given  terms  is 
of  the  same  kind  or  name  as  the  required  answer,  and  place 
it  above  a  horizontal  line,  towards  the  left.  Notice  again 
whether  the  required  term  must  be  greater  or  less  than  this ;  and, 
if  greater,  place  the  greater  of  the  two  remaining  terms  on  the 
right  of  the  preceding  term,  and  also  above  the  line,  and  the 
less  of  the  two  terms  below  the  line ;  but  if  less,  place  the  less 
of  the  remaining  terms  above  the  line,  and  the  greater  below  it ; 
then  cancel,  multiply,  and  divide  as  before  directed. 

Note  2d. — In  arranging  the  terms  as  directed  for  canceling,  the 
number  placed  first  above  the  line,  is  the  third  term  of  the  pro- 
portion, and  that  standing  on  the  same  side,  on  the  right  of  this, 
is  the  second,  and  the  number  standing  below  the  line,  the  first. 

3.  If  12  horses  consume  42  bushels  of  oats  in  a  given  time, 
how  much  will  20  horses  consume  in  the  same  time  ? 


188  PROPORTION. 

Statement,   -jW~* 

The  answer  required  is  obviously  the  oats  that  20  horses 
would  consume.  It  is  also  evident  that  20  horses  would 
consume  more  that  12.  Hence  the  reason  of  the  above  state- 
ment. 

7.  10 

rri>  i  A     *&•  %& 

i  ne  same  canceled  :  —  —  , 

and  7  X  10  =  70,  the  bushels  required. 

It  will  be  recollected  that  the  numbers  above  the  horizontal 
line  form  a  dividend,  and  those  below  the  line,  a  divisor.  Hence, 
42,  the  number  of  bushels  consumed  by  12  horses,  may  be  re- 
garded as  divided  by  12.  This  division  would  give  the  quan- 
tity of  oats  which  one  horse  would  consume  in  the  given  time, 
and  this  quantity  multiplied  by  20  would  give  the  quantity  20 
horses  would  consume  in  the  same  time. 

4.  If  72  yards  of  cambric  cost  $119.44,  what  will  9  yards 
cost  ? 

If  $1  19.44  be  divided  by  72,  the  quotient  will  be  the  price  of 
one  yard,  and  this  price  multiplied  by  9,  will  be  the  required 
answer. 

Therefore,  1-^. 


Canceled,  j%  5    and  11  9.44-^8  =14.  93,  Ans. 

5.  If  10  shillings  pay  for  20  pounds  of  beef,   how  many 
pounds  may  be  bought  for  5  shillings?     Ans.  10. 

Statement,  —  '~^- 

6.  If  3  Ib.  of  sugar  cost  4  s.  how  much  will  6  Ib.  cost  ? 

Ans.  8  s. 

Statement,  -L^-. 
o 

7.  If  12  bushels  of  wheat  are  worth  $16,  what  is  the  value 
of  48  bushels  ?     Ans.  $64. 

8.  Sold  12  yards  of  cloth  for  $72  ;  what  is  the  value  of  5 
yards,  at  the  same  rate  ?     Ans.  $30. 

9.  What  is  the  value  of  16  cords  of  wood,  if  48  cords  are 
worth  $120?     Ans.  $40. 

10.  If  16  cords  of  wood  are  worth  $40,  what  is  the  value 
of  48  cords?     Ans.  $120. 


PROPORTION.  189 

11.  What  is  the  value  of  4  gallons  of  wine,  if  108  gallons  of 
the  same  kind  are  worth  $324?     Ans.  $12. 

12.  If  8  yards  of  cloth  cost  $3.20,  how  many  yards  of  the 
same  kind  may  be  bought  for  $38.40  ?     Ans.  96  yards. 

13.  Paid  75  cents  for  7  Ib.  of  sugar  ;  how  many  pounds  of 
the  same  kind  may  be  bought  for  $6  ?     Ans.  56  pounds. 

14.  If  7  men  can  accomplish  a  piece  of  work  in  12  days, 
how  many  men  are  required  to  accomplish  the  same  in  3  days  ? 
Ans.  28  men. 

15.  If  a  ship  sail  24  miles  in  4  hours,  in  how  many  hours 
will  she  sail  150  miles,  if  she  continue   at  the  same  rate? 
Ans.  25  hours. 

16.  If  17  men  perform  a  piece  of  work  in  25  days,  in  how 
many  days  would  5  men  perform  the  same  ?     Ans.  85  days. 

17.  A  staff',  4  feet  long,  casts  a  shadow  6  feet;  another  staff, 
placed  in  the  same  situation,  casts  a  shadow  58  feet ;  what  is 
its  length?     Ans.  38f  feet. 

18.  A  garrison  has  provision  for  8  months,  at  the  rate  of  15 
ounces  per  day;  what  must  be  each  man's  daily  allowance,  in 
order  that  the  same  provision  may  last  them   1 1    months  ? 
Ans.  lO^J  ounces. 

19.  When  a  quarter  of  wheat  affords  60  ten-penny  loaves, 
how  many  eight-penny  loaves  may  be  made  from  the  same  ? 
Ans.  75. 

20.  If  $10  worth  of  provision  serve  7  men  4  days,  how 
many  days  will  the  same  provision   serve  9  men  ?     Ans.  3^- 
days. 

21.  If  12  gallons  of  wine  cost  $30,  what  is  the  value  of  56 
gallons,  at  the  same  price  per  gallon  ?     Ans.  $140. 

22.  If  15  pounds  of  sugar  cost   $1.20,  how  many  pounds 
may  be  bought  for  $38  ?     Ans.  475  pounds. 

23.  If  a  staff,  4  feet  long,  casts  a  shadow  7  feet  long,  what 
is  the  height   of  a  steeple,  whose   shadow  at  the  same  time 
measures  1 98  feet  ?     Ans.  1 1 3|  feet. 

24.  If  a  pole,  6  feet  long,  casts  a  shadow  10  feet  on  level 
ground,  what  would  be  the  length  of  a  shadow  from  a  steeple 
72  feet  high,  at  the  same  time  ?     Ans.  120  feet. 

25.  If  12  men  build  a  house  in  48  days,  in  how  many  days 
can  36  men  do  the  same  work?     Ans.  16  days. 

26.  If  100  men  can  finish  a  piece  of  work  in  12  days,  how 
many  men  will  be  required  to   do  the   same  in  4  days  ?  Ans. 
300  men. 

27.  How  many  men  must  be    employed  to  complete  in  15 
days  what  5  men  can  do  in  24  days  ?     Ans.  8  men. 


190  PROPORTION. 

28.  If  a  man  perform  a  journey  in  3  days,  when  the  days 
are  1 6  hours  long,  how  many  days  of  1 2  hours  each,  will  he 
require  to  perform  the   same,  if  he  continue  to  travel  at  the 
same  rate  ?     Ans.  4  days. 

29.  If  48  men  can  build  a  wall  in  24  days,  how  many  men 
can  do  the  same  in  192  days  ?     Ans.  6  men. 

30.  In  how  many  days  will  8  men  finish  a  piece  of  work 
which  5  men  can  do  in  24  days  ?     Ans.  15  days. 

31.  In  what  time  will  $600  gain  $50  interest,  if  $80  gain 
it  in  15  years  ?     Ans.  2  years. 

32.  When  $100  principal  will  gain  $6  in  12  months,  what 
principal  will  gain  the  same  in  8  months  ?     Ans.  $150. 

33.  How  many  yards  of  cloth  3  qr.   wide,  are  equal  to  3,0 
yards,  5  qr.  wide  ?     Ans.  50  yards. 

34.  How  many  yards  of  paper  l£  yard  wide,  will  be  suffi- 
cient to  hang  a  room  20  yards  square,  and  4  yards  high?    Ans. 
256  yards. 

35.  If  a  board  be  9  inches  wide,  how  much  in  length  will 
make  a  square  foot  ?     Ans.  16  inches. 

36.  What  quantity  of  shalloon  3qr.  wide,  will  line  7|  yards 
of  cloth,  1-|  yards  wide  ?     Ans.  15  yards. 

37.  Lent  a  friend  $100  for  6  months  ;  afterwards  he  lent  me 
$75  ;  how  long  ought  I  to  retain  it  to  balance  my  favor,  allow- 
ing to  each  the  same  rate  per   cent,  of  interest  ?     Ans.  8 
months. 

38.  In  what  time  will  $858  gain  as  much  as  $286  will  gain 
in  12  months  ?     Ans.  4  months. 

39.  If  375  cwt.  may  be  carried  660  miles  for  a  given  sum, 
how  many  cwt.  may  be  carried  60  miles  for  the  same  money  1 
Ans.  4125. 

40.  If  10s.  worth  of  wine  will  suffice  for  46  men,  when  a 
tun  is  worth  $240,  for  how  many  will  the  same  10s.  suffice, 
when  a  tun  costs  $160  ?     Ans.  69  men. 

41.  If  5£  yards  of  muslin  that  is  H  yards  wide,  will  make 
a  dress,  how  many  yards  of  lining  will  be  required,  that  is  but 
3  qr.  wide  ?     Ans.  1 1  yards. 

42.  If  40  rods  in  length  and  4  in  breadth  make  an  acre, 
what  is  the   width  of  a  piece  of  ground  containing  the  same 
quantity,  that  is  24  rods  in  length  1     Ans.  6  rods,  3  yards,  2 
feet. 

43.  An  insurance  company,  consisting  of  82  persons,  sus- 
tains a  loss,  of  which  each  man's  share  was  $12  ;  wh;i.t  would 
their  shares  have  been,  had  the  company  consisted  of  only  :}-2 
persons?     Ans.  $30.75. 


PROPORTION.  191 

44.  If  a  hogshead  of  wine  which  cost  $180,  afford  a  hand- 
some profit,  when  retailed  at  $4  per  gallon,  how  must  another 
be  retailed,  which  cost  $196,  to  gain  the  same  percent.? 
Ans.  $4.355.+ 

45.  A  lot  of  ground  was  walled  in  by   16  men  in  6  days  ; 
the  same  being  demolished,  is  required  to  be  rebuilt  in  4  days  ; 
how  many  men  must  be  employed  ?     Ans.  24  men. 

46.  A  person  by  traveling   12  hours  per  day,  performs  a 
journey  of  800  miles  in  32  days  ;   how  many  days  will  he  re- 
quire to  perform  the  same  journey,  if  he  travel  15  hours  per 
day?     A?is.  25  J  days. 

47.  If  1800  cwt.  may  be  carried  64  miles  for  a  given  sum, 
how  far  may  225  cwt.  be  carried  for  the  same  money?     Ans. 
512  miles. 

48.  If  50  gallons  of  water  fall  into  a  cistern  of  sufficient 
capacity  to  contain  230  gallons,  in  one  hour,  and  by  a  pipe  35 
gallons  be  drawn  off  in  the  same  time,  how  long  will  it  take  to 
fill  the  cistern  ?     Ans.  15  hours,  20  minutes. 

49.  If  150  Ib.  of  soap  cost  $15.60,  what  would  15  Ib.  cost  ? 
Ans.  $1.56. 

50.  How  many  yards  of  cloth  3  qr.  wide   are  equal  to  39 
yards  5  qr.  wide  ?     Ans.  65. 

51.  A  cistern  has  a  pipe  by  which  it  may  be  emptied  in  10 
hours  ;  how  many  pipes  of  the  same  capacity  will  empty  it  in 
30  minutes  ?     Ans.  20  pipes. 

It  has  already  been  said,  that  if  the  given  terms  (see  Note 
1st)  are  of  different  denominations,  they  must  be  reduced  be- 
fore stating  the  sum.  This  labor  is,  for  the  most  part,  saved, 
whenever  the  question  is  solved  by  canceling.  Take  the  fol- 
lowing sums,  in  which  it  is  required  to  find  the  value  of  a 
quantity  in  one  denomination,  the  price  of  some  other  denom- 
ination being  given. 

RULE. —  Write  the  quantity  the  price  of  which  is  required, 
above  a  horizontal  line ;  then  (if  the  price  of  a  lower  denomina- 
tion be  given)  on  the  right  of  this,  also  above  the  line,  place  the 
numbers  required  to  reduce  the  given  quantity  to  that  denomina- 
tion, together  with  the  price  of  the  same  denomination  ;  then  be- 
low the  line,  write  such  numbers  as  will  reduce  the  given  price 
to  the  required  denomination.  But  if  the  price  of  a  higher  de- 
nomination be  given,  and  that  of  a  lower  denomination  be  re- 
quired, place  the  quantity  the  price  of  which  is  required,  as 


192  PROPORTION. 

before,  and  write  the  numbers  necessary  to  reduce  that  quantity 
to  the  denomination  of  which  the  price  is  given,  below  the  line  ; 
then,  lastly,  place  the  quantity  the  price  of  which  is  given  below, 
and  its  price  above  the  line.  Solve  the  statement  by  canceling. 

Note  3d. — If  either  the  given  quantity  or  price  be  of  differ- 
ent denominations,  they  may  be  reduced  to  the  lowest  given, 
before  stating  ;  or,  if  preferred,  the  lower  denominations  may 
be  reduced  to  a  decimal. 

Ex.  1.  How  many  pounds  sterling  will  3  cwt.  of  sugar  cost, 
at  20  pence  per  pound  ? 

The  price  of  3  cwt.  is  required,  and  that  of  one  pound  is 
given.  The  given  price  is  also  in  pence,  and  pounds  sterlino- 

.     ,              '        3.  4.  28.  20    .     , 
are  required.     Hence, i2~20'  1S         required  statement. 

The  numbers  4  and  28  above  the  line,  are  required  to  reduce 
the  3  cwt.  to  pounds,  and  these  pounds  multiplied  by  20,  will 
give  the  price  of  the  whole  quantity  in  pence.  If,  then,  these 
pence  be  divided  by  12  and  20,  they  will  be  reduced  to  pounds 
sterling. 

„.   .                 i      i    S.  4.  28.  2ft 
Statement  solved, — -. 

Therefore,  28  £.  is  the  required  answer. 

2.  How  many  pounds  sterling  will  3  pipes  of  wine  cost,  at 
10  s.  a  gallon  ? 

.    3.  2.  63.  10 
Statement,   ^ — . 

In  this  statement,  the  3  and  63  above  the  line  are  the  numbers 
required  to  reduce  pipes  to  gallons  ;  then,  the  gallons  multiplied 
by  1 0,  will  give  the  cost  in  shillings  ;  and,  lastly,  the  shillings 
divided  by  20,  (the  number  below  the  line,)  will  be  reduced  to 
pounds. 

Statement  canceled,  3-  2-  g.  1^  63  X  3  =  189  £.  Ans. 

3.  How  many  dollars,  New  York  currency,  will  12  cwt. 
of  sugar  cost,  at  lOd.  a  pound?     Ans.  $140. 

Statement,  —     ^   ^ — .    (8  s.  =  $  1 ,  New  York  currency. ) 

4.  How  many  dollars,  New  England  currency,  will  2  hogs- 
heads of  wine  cost, at  6d.  a  pint? 


PROPORTION.  193 

Statement,  2"  63'^'  2Q6.  Ans.  $84.  ($6s.  =  l  N.  E. currency.) 

5.  At  15  pence  a  pound,  what  will  1  cwt.  of  loaf  sugar  cost 
in  dollars,  New  England  currency  ?     Ans.  $23.333.+ 

If  it  is  preferred  to  solve  sums  of  this  kind  without  cancel- 
ing, it  may  be  done  by  the  following  rule  : 

RULE. — Reduce  the  given  quantity  to  that  denomination,  the 
price  of  which  is  given,  and  multiply  it  by  the  price ;  then  divide 
by  such  numbers  as  are  required  to  reduce  the  value  obtained  to 
the  required  denomination. 

6.  How  many  dollars,  New  York  currency,  will  6  cwt.  of 
sugar  cost  at  1 0  d.  a  pound  ? 

SOLUTION. 
6 
4=qr.  in  cwt. 

2  4 

2  8=:lbs.in  qr. 

1  92 

48 

672 

1  0=price  of  1  Ib. 

12)672  Oncost  in  pence. 


8)56  Oncost  in  shillings. 

7  Orr  dollars,  answer. 
7 
Canceled,  —H^.  10x7=70,  answer. 

135.     a. 
S     X 

7.  How  many  dollar^  will  53  ells  English  cost,  at  8  s.  New 
York  currency,  per  yard  ?     Ans.  $66.25. 

8.  If  I  purchase  melasses  at  1  s.  3  d.  per  quart,  how  much  in 
pounds,  shillings,  and  pence,  will  12  hogsheads  of  the  same 
kind  cost?     Ans.  189£. 

9.  If  I  purchase  16  cwt.  of  steel  for  $156,  what  will   1  qr. 
of  a  cwt.  cost,  at  the  same  rate  ? 

17 


194  PROPORTION. 

39 

Statement,  i^yjl.    Canceled,^*;  39 -f- 16  =  $2. 437.+ 

10.  What  cost  9  cwt.  of  sugar  at  10  pence  per  pound  ?  Ans. 
42  £. 

11.  What  cost  12  cwt.  of  sugar  at  9  pence  per  pound  ?  Ans. 
50  £.  8s. 

12.  What   cost  42  cwt.  of  sugar  at  3s.  8d.  per  pound? 
Ans.  862  £.  8  s. 

13.  What  would  480  yards  cost,  in  federal  money,  at  2  pence, 
New  York  currency,  per  yard  ? 

480     2 
Statement,  —  '  ^  Q.     Ans.  $10.+ 

For  the  solution  of  the  following  sums,  see  the  table  of  Cur- 
rencies, given  in  Reduction  of  Currencies. 

14.  What  would  862  yards  cost,  in  federal  money,  at  3  pence 
per  yard,  New  England  currency  1 

8fi2    *} 
Statement,  --J£-Q>     Ans.  $35.916.+ 

15.  What  would  920  yards  cost,  in  federal  money,  at  4^ 
pence  per  yard,  New  Jersey  currency  ?     Ans.  $46. 

16.  What  would  988  pounds  of  rice  cost,  in  federal  money,  at 
6  d.  per  pound,  South  Carolina  currency  ?     Ans.  $105.857.+ 

17.  What  would  899   gallons  of  vinegar  cost,   in  federal 
money,  at  8  d.  per  gallon,  New  York  currency  ?     Ans.  $74.- 
916.+ 

18.  How  much  will  672  yards  cost,  in  federal  money,  at  6s. 
6  d.  New  York  currency,  per  yard  ?     Ans.  $546. 

19.  What  will  1000  yards  of  ribbon  cost,  in  federal  money, 
at  3  s.  4  d.  New  England  currency,  per  yard  ?     Ans.  $555.- 
555.+ 

20.  How  many  dollars  will  123  yards  of  cloth  cost,  at  10s. 
New  Jersey  currency,  per  yard  ?     Ans.  $164. 

21.  What  will  687  yards  of  cloth  cost  in  federal  money,  at 
5  s.  South  Carolina  currency,  per  yard  ?     Ans.  $736.071.+ 

22.  How  many  dollars  and  cents  will  127  gallons  of  wine 
cost,  at  3  s.  4  d.  New  England  currency,  per  gallon  ?     Ans. 
$70.555.+ 

23.  How  many  dollars  will  pay  for   14  cwt.  3qr.  of  hay, 
at  15  s.  8  d.  New  York  currency,  per  cwt.  ?     Ans.  $28.885.  + 

24.  How  many  dollars  will  pay  for  12  cwt.  2  qr.  of  cheese, 
at  2.C.  15s.  New  Jersey  currency,  per  cwt.?   Ans.  $91.666.+ 


PROPORTION.  195 

25.  What  will  12  cwt.  7  Ib.  of  brown  sugar  cost,  at  6d.  New 
Jersey  currency,  per  pound  1     Ans.  $90. 066. -f- 

26.  A.  owes  B.  1 138  £.  but  can  pay  only  15  s.  on  a  pound  ; 
what  will  B.  receive  ?     Ans.  853  £.  10s. 

27.  C.  has.  a  journey  of  75  leagues  to  perform.     In  what 
time  will  he  complete  it,  if  he  travel  30  miles  a  day.     Ans. 
7|-  days. 

28.  How  long  will  it  take  to  travel  one  fourth  round  the 
earth,  at  the  rate  of  36  miles  a  day  ;  the  whole  circumference 
being  360  degrees,  and  one  degree  69|   miles  ?     Ans.    173| 
days. 

29.  If  9  Ib.  of  coffee  cost  27  s.  how  many  dollars,  at  6s. 
each,  will  45  Ib.  cost  ?     Ans.  $22.50. 

30.  If  I  buy  20  pieces  of  cloth,  each  20  ells,  for  12  s.  6  d. 
per  ell,  how  many  dollars,  at  8  s.  will  pay  for  the  same  1    Ans. 
$625. 

31.  How  many  dollars  will  7  casks  of  prunes  cost,  each 
weighing  4  cwt.  2  qr.,  at  2  £.  19  s.  8  d.  New  York  currency,, 
per  cwt.  ?     Ans.  $234.937.  + 

32 .  A  vessel  at  sea  discharges  a  cannon,  the  report  of  which 
reaches  me  in  1  minute,  30  seconds.     How  far  distant  is  she, 
allowing  sound  to  travel   1142  feet  in  a  second?     Ans.   19 
miles,  3  furlongs,  and  29TJT  rods. 

33.  How  many  yards  of  flannel,  1  yard  wide,  will  line  125 
yards  of  broadcloth,  1  ell  English  wide  ?     Ans.  156^. 

34.  How  many  yards  of  cloth  may  be  bought  for  $37.62,  if 
£  of  a  yard  cost  66  cents  ?     Ans.  $42  yd.  3  qr. 

35.  How  many  dollars  will  28  yards  of  linen  cost,  at  5  s.  6 
d.  New  England  currency,  per  yard  ?     Ans.  $25.666. -f- 

36.  Bought  32  yards  of  muslin,  at  6  s.  8  d.  New  York  cur- 
rency, per  yard.    What  was  the  cost,  in  federal  money  ?    Ans. 
$26.666.+ 

37.  How  many  gallons  of  wine  may  be  purchased  for  $45, 
at  6  s.  New  York  currency,  per  gallon  ?     Ans.  60  gallons. 

QUESTIONS. — What  is  ratio  1  What  is  the  former  of  the  two  num- 
bers between  which  the  ratio  exists  called?  What  is  the  latter7?  How 
is  the  direct  ratio  of  any  two  numbers  obtained  1  How  is  the  inverse 
ratio  obtained  1  Between  what  quantities  only,  does  ratio  exist  1  How 
is  simple  ratio  expressed  1  How  is  the  ratio  of  any  couplet  affected  by 
multiplying  or  dividing  both  the  antecedent  and  the  consequent  by 
the  same. number  1  In  what  two  ways  may  the  ratio  of  any  two  num- 
bers be  multiplied'?  In  what  two  ways  is  the  ratio  of  any  couplet 
divided  by  any  number  1  When  two  or  more  ratios  are  multiplied 
together,  what  is  the  resulting  ratio  called1?  What  does  it  equal? 


196  COMPOUND  PROPORTION. 

What  is  Note  1st  1  What  constitutes  proportion?  How  many  equal 
ratios  are  required  for  a  statement  of  proportion  1  What  terms  must  be 
of  the  same  kind?  When  any  four  terms  are  proportional,  what  are  the 
first  and  fourth  called  1  And  what  are  the  second  and  third  called  1 
How  does  the  product  of  the  extremes  compare  with  the  product  of  the 
means  ?  When  any  of  the  four  terms  are  wanting,  ^explain  how  it 
may  be  found.  In  what  do  operations  in  Simple  Proportion 
consist  1  What  is  the  rule  1  What  note  follows  the  rule  ?  What  is 
the  rule  for  canceling  1  What  is  Note  2d  1  What  is  the  rule,  when  it 
is  required  to  find  the  value  of  a  quantity  in  one  denomination,  the 
price  of  some  other  denomination  being  given  1  What  is  Note  3d  1 
What  is  the  rule  for  solving  sums  of  this  kind  without  canceling  ? 


COMPOUND     PROPORTION. 

Simple  Proportion  consists  of  two  equal  ratios. 

Compound  Proportion  is  that  in  which  the  relation  of  one  of 
the  given  quantities  to  a  required  quantity  of  the  same  name, 
is  traced  through  two  or  more  simple  proportions. 

The  smallest  number  of  terms  of  which  a  statement  in  com- 
pound proportion  can  consist,  is  five.  Of  these  terms,  one  is 
always  of  the  same  name  as  the  answer  required  ;  and  the 
others  are  always  two  of  a  kind.  The  following  sum  will 
serve  as  an  illustration  : 

If  3  men,  in  4  days,  spend  $5,  how  many  dollars  will  6  men 
spend  in  12  days  ? 

In  the  above  sum,  there  are  five  terms  given,  viz.  two  of 
men,  two  of  days,  and  one  of  dollars ;  and  dollars  are  also 
required  for  the  answer ;  so  that  when  the  sixth  term  is  found, 
the  sum  may  be  resolved  into  three  simple  ratios,  the  third  of 
which  is  a  compound  of  the  preceding  two.  These  ratios  are 
3  men :  6  men,  and  4  days :  12  days,  and  $5  :  the  required  number 
of  dollars.  Now  it  is  obvious,  that  the  ratio  of  $5  to  the  required 
number  of  dollars,  is  not  the  same  as  the  ratio  of  3  men  to  6  men ; 
for  then  no  regard  would  be  paid  to  the  time,  and  the  solution 
would  be  effected  on  the  supposition  that  one  man  or  a 
number  of  men  would  spend  as  much  in  one  day,  as  in  any 
given  number  of  days.  Nor  is  it  the  same  as  the  ratio  of  4 


COMPOUND  PROPORTION.  197 

days  to  12  days,  for  then  the  supposition  would  be,  that  3  men 
would  spend  as  much  as  6,  or  any  number  of  men.  But  it  is 
a  ratio  compounded  of  the  two  ratios,  viz.  the  ratios  of  3  men 
:  6  men,  and  of  4  days  :  12  days.  The  ratio  of  3  :  6=2,  and 
$5  x  2  =  $  1 0.  This  would  double  the  given  quantity  of  money. 
Again,  the  ratio  of  4:  12,  is  3  ;  this  would  treble  the  sum  last 
obtained,  viz.  $10,  and  would  give  10x3  =  $30,  which  is  the 
answer  to  the  above  question.  Now,  it  will  be  observed,  that 
the  $5  in  the  above  operation  was  multiplied  by  the  product 
of  the  two  simple  ratios  ;  for  3  :  6=2,  and  4  : 12  =  3  ;  there- 
fore, 2  x  3  =  6,  the  product  of  the  simple  ratios,  and  $5x6  = 
$30,  Ans. 

The  same  result  would  have  been  obtained,  by  multiply- 
ing the  $5  by  the  consequents,  or  latter  terms  of  each  ratio, 
and  dividing  their  product  by  the  product  of  the  ante- 
cedents, or  former  terms  of  the  same  ratios.  Thus,  3  :  6  and 
4  :  12,  are  the  given  ratios  of  which  6  and  12  are  consequents; 
therefore,  6  x  12  x  5  =  360,  and  3  X  4  =  12,  the  product  of  the 
antecedents;  hence,  360  — 12  =  30  dollars,  the  same  answer 
as  before.  Hence,  we  have  the  following  rule: 

RULE. — Write  the  number  which  is  of  the  same  kind  as  the 
answer  required,  for  the  third  term.  Of  the  remaining  terms, 
take  any  two  of  the  same  name,  and  arrange  them  as  directed 
in  Simple  Proportion  ;  then  any  other  two  of  a  kind,  and  so  on 
till  the  terms  are  all  taken.  Lastly,  multiply  the  product  of  the 
second  terms  by  the  third  term,  and  divide  the  last  product  by  the 
product  of  the  first  terms,  and  the  quotient  will  be  the  required 
term  or  answer. 

Ex.  1 .  If  4  men  build  a  wall  ten  feet  long,  3  feet  high,  and 
2  feet  thick,  in  6  days,  in  what  time  will  12  men  build  one 
100  feet  long,  4  feet  high,  and  3  feet  thick?  The  question 
asked  is,  in  what  time  will  the  work  be  done  ;  therefore,  by  the 
rule,  6  days  is  the  third  term. 

12  men  :  4  men,  ~| 

10  feet  length  :  100  feet  length,  [ 

3  feet  high:  4  feet  high,        '  f  : '  6  days. 

2  feet  thick :  3  feet  thick,        J 

It  is  obvious  that  12  men  would  require  less  time  to  execute 
a  given  piece  of  work  than  4  men,  and  also  that  a  wall   100 
17* 


198  COMPOUND  PROPORTION. 

feet  long,  4  feet  high,  and  3  feet  thick,  would  require  more  time 
than  a  wall  10  feet  long,  3  feet  high,  and  2  feet  thick. 

The  same  solved  : 
12:4 

;  4x100x4x3x6=28800;  and  12x10x3 
X  2=720  ;  and  28800-4-720=40  days,  the 
number  required. 

The  work  of  this  sum  maybe  much  abbreviated  by  canceling. 
The  statement  would  be,  6"  *'  *•  *'  * 


1.6.      1U.    O.    xJ 

Canceled,  ?i^A|i|,   10x4=40, 

For  the  preceding  mode  of  operation,  we  have  the  following 
rule  : 

RULE  FOR  CANCELING.  —  Write  the  number,  which  is  of  the 
same  name  as  the  answer  required,  above  a  horizontal  line, 
towards  the  left.  Then  take  each  two  terms  of  the  same  name, 
and  compare  them  with  this  number,  and  notice  whether  more  or 
less  be  required,  as  in  Simple  Proportion  :  if  more  be  required, 
place  the  greater  of  the  two  numbers  above  the  line,  and  the 
less  below  ;  but  if  less  be  required,  place  the  less  of  the  two 
above  the  line,  and  the  greater  below.  Cancel,  fyc. 

Ex.  2.  If  4  men  in  12  days  mow  48  acres  of  grass,  how 
many  acres  will  8  men  mow  in  1  6  days  1 

A    •  Q      ^ 

Common  statement,       ;  '       v  :  ;  48. 

Statement  for  canceling,  48'  S'  *|*. 
4    2 

£Q       "G       1  £ 

Canceled,  —  L-^1-^'  4x2  x  16  =  128,  the  number  of  acres* 
required. 

Note  1st.  —  If  the  sixth  term,  or  answer,  be  placed  below 
the  horizontal  line,  the  other  term  remaining  as  before,  the 
product  of  the  terms  standing  above  the  line,  will  just  equal 
the  product  of  those  standing  below,  thus  : 
48X8X16=6144 
128X4X12=6144' 


PROPORTION.  199 

3.  If  12  horses  eat  20  bushels  of  oats  in   16  days,  how 
many  bushels  will  24  horses  eat  in  48  days  ? 

Statement,  —  -y!^|     Ans.  120  bushels. 

4.  If  18  men,  in  32  days,  consume  128  Ibs.  of  bread,  how 
many  pounds  will  12  men  consume  in  64  days,  their  daily 
allowance  being  the  same  ? 

1 0Q      1 0     RA 

Statement,          \'  "v     A™.  170}. 
ib.  0*4 

5.  If  8  men  receive  4  dollars  for  3  days  work,  how  many 
days  must  20  men  work,  to  earn  40  dollars,  if  they  receive 
the  same  daily  allowance  ?     Ans.  12. 

Statement,         ' — -. 

To  understand  this  statement,  it  should  be  remembered,  that 
20  men  would  require  less  time  for  the  accomplishment  of  an 
object,  than  8  men ;  and  also  that  40  dollars  would  employ, 
for  a  given  time,  more  men  than  4  dollars. 

6.  If  4    men  receive  $3.20  for  3  days  work,  how  many 
men,  if  they  receive  the  same  per  day,  will  eam  $12.80,  in 
16  days  ?     Ans.  3  men. 

4.  12.80.  3 
btatement,  — Too~Tfi" 

7.  How  much  ought  60  men  to  receive  for  25  days  work, 
if  12  men,  under  the  same  circumstances,  receive  50  dollars 
for  4  days  work  ?     Ans.  $1562.50. 

8.  If  16  men  cut  112  cords  of  wood  in  7  days,  how  many 
cords  will  24  men  cut  in  19  days  ?     Ans.  456  cords. 

9.  If  the  transportation  of  12  cwt.  150  miles,  cost  75  shil- 
lings,  for  how  many  dollars,    at  8  s.  each,  may  6  cwt.  be 
transported  45  miles  ?     Ans.  $1.406.+ 

10.  If  the  freight  of  9  hhds.  of  sugar,  each  weighing  12 
cwt.  20  leagues,  cost  $38.40,  what  will  be  the  expense  of 
transporting  50  casks  of  the   same,  each  weighing  2  cwt.  2 
qr.,  100  leagues?     Ans.  $222.22.+ 

11.  If  100  dollars,  in  1  year,  gain  6  dollars  interest,  how 
much  will  200  dollars  gain  in  26  weeks.     Ans.  6  dollars. 

12.  If  350    dollars,  in  9  mdtaths,  gain  15  dollars,   what 
principal  will  gain  6  dollars  in  12*nonths  ?     Ans.  105  dollars. 

13.  If  8  men  can  build  a  wall,"2£)  feet  long,  6  feet  high,  and 
4  feet  thick,  in  12  days,  in  what  time  can  24  men  build  one, 
200  feet  long,  8  feet  high,  and  6  feet  thick  ?     Ans.  80  days. 

14.  A  wall,  32  feet  high,  and  40  feet  long,  was  built  in  8 


200  PROPORTION. 

days,  by  145  men;  in  how  many  days  would  68  men  build 
another  wall,  28  feet  high,  and  of  the  same  length,  allowing 
each  man  to  perform  an  equal  portion  of  labor,  in  the  same 
time?  Ans.  14||- days. 

15.  What  must  be  paid  for  the  transportation  of  56  bags  of 
coffee,  each  weighing   3  qr.   16  lb.,   66  miles,    if  14  bags, 
weighing  each  125  lb.,  be  carried  6  miles  for  $6.25.     Ans. 
220  dollars. 

16.  If  6  persons  can  earn  120  £.  in  21  weeks,  how  much 
will  14  persons  earn,  in  46  weeks,    if  they  receive  the  same 
per  week?     Ans.  613  £.  6  s.  8  d. 

17.  If,  when  the  days  are  14  hours  long,  a  person  perform  a 
journey  of  276  miles  in  16  days,  in  what  time  will  he  travel 
860  miles,  when  the  days   are  12  hours  long  ?     Ans.  58^4T 
days. 

18.  If  960  dollars  defray  the    expenses  of    20    men   88 
weeks,  for  how  many  weeks  will  $1440  defray  the  expenses 
of  48  men,  if  they  spend  at  the  same  rate  ?     Ans.  55  weeks. 

19.  If  100  jC.  gain  6  £.  in  12  months,  what  principal,  at  the 
same   rate  per   cent.,  will  gain  3  £.  7s.  6  d.  in  9  months  ? 
Ans.  75  £. 

20.  If  a  man  travel  240  miles  in  12  days,  when  the  days 
are  12  hours  long,  in  how  many  days,  of  16  hours,  will  he 
travel  720  miles,  if  he  travel  at  the  same  rate  ?     Ans.  27  days. 

21.  What  is  the  interest  of  $650  for  36  weeks,  at  6  per 
cent,  per  annum  ?     A?is.  $27. 

22.  If  $150,    in  12  months,  gain  $8  interest,  what  will 
$400  gain  in  4  months  ?     Ans.  $7.11.+ 

23.  If  3  men  lay  144  square  yards  of  pavement  in  7  days, 
how  many  square  yards  will  12  men  lay  in  49  days  ?       Ans. 
4032  square  yards. 

24.  If  200  lb.  be  carried  40  miles  for  40  cents,  how  far 
may  202000  lb.  be  carried  for  $60.60  ?     Ans.  60  miles. 

25.  Inwhattime  will200£.  gain  6£.  if  100£.  gain  6£.  in 
52  weeks  ?     Ans.  26  weeks. 

26.  In  what  time  will  400  £.  gain  96  £.  interest,  if  350  £. 
gain  10  £.  10s.  interest  in  6  months  ?     Ans.  4  years. 

27.  If  4  men  mow  48  acres  of  grass   in  12  days,  in  what 
time  will  8  men  mow  128  acres  ?     Ans.  16  days. 

28.  If  I  receive  88£.  17  s.  4  d.  for  the  principal  and  int< 

of  86  £.  for  8  months,  what  is  the  rate  per  cent,  of  interest  ? 
Ans.  5  per  cent. 

29.  What  will  the  transportation  of  7  cwt.  2  qr.  25  lb.  64 


PROPORTION.  201 

miles  cost,  if  5  cwt.  3  qr.  be  carried  150  miles  for  $24.58  ? 
Ans.  $14.086.+ 

30.  If  I  pay  $50.25  for  the  tuition  of  2  boys,  3  quarters  each, 
what  will  the  tuition  of  100  boys  amount  to,  in  7£  years,  at 
the  same  rate?     Ans.  $25125. 

31.  Purchased  goods  to  the  amount  of  750  £.  and  sold  the 
same,  six  months  after,  for  825  £. ;  what  did  I  gain  per  cent.  ? 
Ans.  20  per  cent. 

32.  If  56  Ib.  of  bread  be  sufficient  for  7  men   14  days, 
how  long  will  36  Ib.  suffice  for  21  men  ?     Ans.  3  days. 

33.  A  person  having  engaged  to  carry  8000  cwt.  a  certain 
distance  in  9   days,  removed  4500  cwt.   with  18  horses,  in 
6   days  ;  how  many  horses  were  required   to  remove  the  re- 
mainder, in  the  3  remaining  days  ?     Ans.  28  horses. 

34.  If  3   men  perform  a  piece  of  work  in  20  days,  how 
many  men  will  accomplish  4  times  as  much  work  in  4  days  ? 
Ans.  60  men. 

35.  If  4  men,  in  5  days,   eat  6  Ib.  of  bread,  how  many 
pounds  of  the  same   will  be   sufficient  for   16  men  15  days? 
Ans.  72  Ib. 

36.  If  it  take  15  men  20  days  to  make  300  pairs  of  shoes, 
how  many  men  will  be^  required  to  make  1200  pairs  in  60 
days  ?     Ans.  20  men. 

37.  A  wall,  which  was  to  be  raised  to  the  height  of  27  feet, 
was  raised  9  feet  by  12  men,  in  6  days  ;  how  many  men  were 
required  to  complete  the  work  in  4  days,  allowing,  each  man 
to  do  the  same  amount  of  work  per  day  ?     Ans.  36  men. 

38.  If  6  men,  in  21  weeks,  earn  $120,  how  much  will  14 
men  earn  in  46  weeks  ?     Ans.  $613.33.+ 

39.  If   48  bushels  of  corn  produce  576  bushels    in  one 
year,  what  will  be   the  product  of  240  bushels  in  6  succes- 
sive years?     Ans.  17280  bushels. 

40.  In  how  many  days  will  25  men  reap  200  acres  of  grain, 
if  12  men  reap  80  acres  in  6  days  ?     Ans.  7J, 

41.  If  20  men  mow  208  acres,  1  rood,  and  24  rods  of  grass, 
in  24  days,  how  many  men  will  cut  down  8  times  as  much 
grass  in  twice  the  time  ?     Ans.  80. 

QUESTIONS. — Of  what  does  simple  proportion  consist?  What  is 
compound  proportion'?  What  is  the  smallest  number  of  terms  of  which 
a  statement  in  compound  proportion  can  consist  1  How  must  these 
terms  compare  with  each  other,  in  kind?  How  many  simple  ratios 
will  there  be  in  the  example  given  for  illustration,  when  the  sixth  term 
is  found  1  Of  what  is  the  third  ratio  a  compound  1  What  are  the  three 
ratios  in  the  illustration  given?  Give  the  entire  illustration.  What  is  the 


202  ANALYTICAL  SOLUTIONS. 

rule  first  given  7  What  is  to  be  the  denomination  of  the  third  term  7 
How  are  the  other  terms  to  be  arranged?  What  is  the  rule  for  can- 
celing 1  What  term  is  to  be  placed  first  above  the  line  1  How  are  the 
other  terms  to  be  arranged  1  What  is  the  given  note  1 


SOLUTION  OF  ARITHMETICAL 
PROBLEMS  BY  ANALYSIS. 

An  arithmetical  question  is  solved  analytically,  when  the 
operation  is  guided  entirely  by  the  conditions  embraced  in  the 
question  itself. 

Take  the  following  illustration  : 

Ex.  1.  If  3  men  perform  apiece  of  work  in  6  days,  in  what 
time  will  9  men  perform  the  same  labor  ? 

It  is  obvious  that  if  it  take  3  men  6  days  to  perform  the 
proposed  labor,  it  will  take  one  man  3  times  6,  or  18  days,  to 
perform  the  same.  But  9  men  operating  together,  will  per- 
form 9  days  work  in  one  day ;  consequently,  they  will  do  the 
whole  in  2  days  ;  for  18-^9=2,  Ans. 

2.  If  21  men  earn  63  dollars  in  a  given  time,  how  much 
will  42  men  earn  in  the  same  time  ? 

63-^21  =3,  the  number  of  dollars  one  man  will  earn  in  the 
given  time.  Therefore,  $42  x  ^—  $126,  the  answer  required. 
Or  the  ratio  of  21  men-to  42  is  2  ;  and,  $63x2=r$126,  the 
same  as  before. 

Solutions  of  this  kind  may,  therefore,  be  effected  by  the 
following  general  principle  : — Find  the  ratio  of  the  two  given 
terms  which  are  of  the  same  kind,  and  by  this  ratio  multiply 
the  term  corresponding  in  kind  with  the  one  required. 

3.  If  42  men  can  make  3  rods  of  wall  in  a  given  time, 
how  much  can  8  men  make  in  the  same  time  ? 

The  ratio  of  42  men  to  8  men  is  42  :  8=-f^=-^j  ;  therefore, 
Jy  X  3=Jf =j  of  one  rod,  which  is  the  distance  required. 

4.  If  a  staff  4  feet  long,  cast  a  shadow  6  feet,  how  high 
is  that  steeple,  whose  shadow  measures  75  feet  ? 


ANALYTICAL   SOLUTIONS.  203 

The  ratio  of  6  :  75  =  12£,  and  12£  x4=50  feet,  Ans.  Or, 
the  shadow  is  1^=§  as  long  as  the  staff;  hence,  75-^3  = 
25,  and  25  X  2  =  50  feet,  the  same  answer  as  before. 

5.  An  express  traveling  at  the  rate  of  60  miles  per  day,  had 
been  absent  5  days,  when  a  second  express  was  dispatched  on 
the  same  rout,  traveling  75  miles  per  day.     How  many  miles 
must  the  second  travel  to  overtake  the  first  ? 

60x5  =  300,  the  whole  number  of  miles  traveled  by  the 
first  express  before  the  second  started,  and  consequently,  the 
number  of  miles  the  second  had  to  gain.  But  the  first  travels 
60,  and  the  second  75  miles  per  day  ;  hence,  75 — 60  =  15,  the 
number  of  miles  gained  daily,  by  the  second  express.  15 
miles  are,  therefore,  gained  in  traveling  75  miles,  consequent- 
ly, one  mile  is  gained  in  traveling  5  miles  ;  and  since  300  miles 
are  to  be  gained,  300x5  =  1500  miles,  answer. 

6.  If  6  men  in  14  days  earn  84  dollars,  how  much  will  9 
men  earn  in  1 1  days  ? 

$84 -H  6  =  $14,  the  money  one  man  will  earn  in  14  days,  and 
$14-^14  =  $!,  the  wages  of  one  man  for  one  day;  therefore, 
$1  X  9  =  $9,  the  money  9  men  will  earn  in  one  day,  and  $9  X 
11  =  $99,  the  money  9  men  will  earn  in  11  days. 

7.  If  6  persons  spend  $300  in  8  months,  how  much  will  be 
sufficient  for  a  family  of  1 5  persons  20  months  ? 

300-^-6  =  50,  and  50^-8  =  $6|,  the  money  spent  by  one  per- 
son in  one  month  ;  then,  $6£  X  15  X  20  =  $1 875,  answer. 

8.  If  12  men  build  36  feet  of  wall  in  9  days,  how  many 
men  would  build  108  feet  in  16  days  ? 

36-1-12  =  3,  and  3-^-9=^,  the  distance  built  by  one  man  in 
one  day  ;  and  ^  x  16  =  ^=5^,  the  distance  one  man  would 
build  in  16  days;  therefore,  108-^5|-=20£,  the  number  of 
men  required. 

9.  A  merchant  owning  |-  of  a  vessel,  sold  J  of  his  share  for 
$1456.     What  was  the  value  of  the  whole  vessel? 

|  of  f  =  &.  If  then,  T85  cost  $1456,  1456^8  =  $182,  the 
value  of  -J-L  of  the  vessel ;  hence,  182  x  15  =  $2730,  answer. 

10.  If  J  of  a  yard  cost  £  of  a  dollar,  what  will  40  yards  cost  ? 
If  f  of  a  yard  cost  $J,  J  of  a  yard  will  cost  ^  of  that  sum, 

or  $•  of  $^=$^j,  and  one  yard  or  J  will  cost  5  times  that 
sum,  or  $ff ;  therefore,  $|f  X  40  =  $1-J£°  =  $58.333,+  Ans. 

11.  If  240  men  perform  a  piece  of  work  in  8  months,  how 
many  men  must  be  employed  to  finish  the  same  work  in  2 
months  ? 

The  ratio  of  2  :  8=4,  and  240  X  4=960  men,  answer. 


204  ANALYTICAL    SOLUTIONS. 

APPLICATION  OF  CANCELING  TO  ANALYTICAL  SOLUTIONS. 

12.  If  8  pounds  of  tea  cost  $12,  what  will  32  pounds  cost  ? 

12    32 

Statement,    — — -=.. 
o 

The  two  terms  of  the  same  name  here  given  are  8  and  32, 
and  their  ratio  is  4,  and  is  obtained  by 

4 

12   &J 
Canceling,  — '- — -. 

Therefore,  $12x4  =  $48,  answer.  The  third  term  is  here 
multiplied  by  the  ratio  of  the  first  and  second,  as  required  for 
analytical  solution.  The  terms  are  also  canceled  and  multi- 
plied as  directed  by  the  rule  for  canceling. 

13.  If  16  horses  consume  84  bushels  of  grain  in  24  days, 
how  many  bushels  will  suffice  32  horses  48  days  ? 

In  the  preceding  sum,  it  is  evident  that  the  given  quantity  of 
grain  is  to  be  increased  by  the  ratios  of  16  horses  to  32  horses, 
and  of  24  days  to  48  days.  Hence, 

Q  A       oo       A  Q 

—     '    '  is  the  statement  expressing  those  ratios. 

2     2 

Canceled,  —  ^'  ^,  84x2x2=336, 
the  number  of  bushels  required. 

We  therefore  see  by  the  above  example,  that  the  effect  of  the 
operation  is  to  increase  the  quantity  of  the  same  name  as  the 
required  quantity,  by  all  the  given  ratios.  The  same  is  true  in 
all  cases,  that  is,  every  statement  for  canceling  is  a  complete 
analysis  of  the  question  under  consideration. 

14.  If  8  men  build  9  feet  of  wall  in  1 2  days,  how  many  men 
must  be  employed  to  build  36  feet  in  4  days  ? 

8.  36.  12 
statement, — -?• 

The  number  of  men  required  will  obviously  depend  on  the 
the  ratios  9 :  36,  and  4:12,  the  former  of  which  is  4,  and  the 
latter,  3.  Therefore,  8  men  x  4  X  3  =  96  men,  the  number  re- 
quired. The  above  statement  canceled,  gives  the  same  result, 
thus: 

4     3 

r-^?,  8X4X3  =  96  men. 


ANALYTICAL  SOLUTIONS.  205 

The  same  number  would  have  been  obtained,  had  the  num- 
bers been  canceled  in  any  other  order  ;  thus, 

*L^J|,  and  8x12  =  96. 
s.    ^* 

Hence  we  perceive,  that  in  a  correct  solution  of  any  sum 
by  canceling,  a  complete  analysis  of  that  sum  is  given. 

15.  If  10  men  make  300  pairs   of  boots  in  20  days,  how 
many  men  must  be  employed  to  make  450  pairs  in  30  days  ? 
Ans.  10  men. 

~    4  k    10.  450.  20 

Statement,  --  m~^. 

If  10  men  make  300  pairs  in  20  days,  they  would  make  15 
pairs  in  one  day  ;  and  if  10  men  make  15  pairs  in  one  day,  one 
man  would  make  one  and  a  half  pairs  per  day  ;  and  in  30  days, 
he  would  make  45  pairs  ;  therefore,  450  -^-45  =  10,  Ans. 

16.  If  £  of  a  yard  cost  f  of  a  pound  sterling,  what  will  f  of  a 
yard  of  the  same  cloth  cost  ? 

If  \  yard  cost  f  of  a  pound,  the  whole  yard  would  cost  J  of 
a  pound,  and  £  of  the  same  would  cost  £  of  J  of  a  pound  =-£$ 
of  a  pound  ;  consequently,  f-  would  cost  5  times  that  sum,  or  |§ 
=  |  of  a  pound,  or  15s.  Ans. 

17.  If  T7g-  of  a  house  cost  49  pounds,  what  will  be  the  value 

of  ^  of  tne  same  ?     Ans-  10  £-  10  s- 

1  8.  A  merchant  bought  a  number  of  bales  of  velvet,  each 
containing  129j-f  yards,  at  the  rate  of  $7  for  5  yards,  and  sold 
the  same  at  the  rate  of  $11  for  7  yards,  and  gained  $200  by 
the  transaction.  How  many  bales  were  there  ? 

He  paid  J-  of  a  dollar  per  yard,  and  received  L1  of  a  dollar 
for  the  same.  Hence,  L1  —  s^ff  —  ft  =  3T  of  one  dollar,  the 
amount  gained  on  one  yard.  Therefore,  $200-f-¥65  =  T-°^-0, 
the  whole  number  of  yards;  and  7-°g0-0^-129i|  —  7-W  — 


19.  If  7  horses  consume  2f  tons  of  hay  in  6  weeks,  how 
many  tons  will  12  horses  consume  in  8  weeks  ?    Ans.  6^  tons. 

20.  If  14  men  finish  a  piece  of  work  in  42  days,  how  long 
will  it  take  21  men  to  do  it  ?     Ans.  28  days. 

21.  If  f  of  a  farm  be  valued  at  $895,  what  is  the  whole 
farm  worth  ?     Ans.  $1611. 

22.  If  7  horses   consume  29  bushels  of  oats  in  5  weeks, 
how  many  will   12  horses  consume  in  6  weeks  ?     Ans.  59Jf 
bushels. 

23.  A  merchant  owning  |  of  a  vessel,  sold  f  of  his  share 

18 


206  PROPORTION  IN  FRACTIONS. 

for  $1200  ;  what  was  the  value  of  the  whole  vessel,  at  the  same 
rate?     Ans.  $1645.714.+ 

24.  There  is  a  pole  £  in  the  mud,  f  in  the  water,  and  8  feet 
out  of  the  water.     What  is  its  length?     Ans.  53^  feet. 

25.  In  a  certain  orchard  -5  of  the  trees  bear  apples,  i  pears, 
i  plums,  30  of  them  peaches,  and  20   cherries.     How  many 
trees  does  the  orchard  contain  ?     Ans.  600. 

26.  A    certain  school  is   classified  as  follows  :   -f^  study 
grammar,  f  study  geography,  -f^   arithmetic,  -^  write,  and  9 
learn  to  read.     How  many  are  there  in  all,  and  how  many  in 
each  study  ?     Ans.  Whole  number  80.     In  grammar  5,  geog- 
raphy 30,  arithmetic  24,  writing  12,  and  9  read. 


SIMPLE  AND  COMPOUND  PROPORTION 
IN  FRACTIONS. 

In  stating  such  sums  in  Simple  or  Compound  Proportion  'as 
consist  of  fractions,  it  is  only  necessary  to  compare  terms  as 
already  directed,  and  then,  if  they  are  solved  without  cancel- 
ing, having  inverted  the  divisor,  to  divide  the  product  of  the  nu- 
merators by  the  product  of  the  denominators.  If,  however, 
they  are  to  be  solved  by  canceling,  arrange  the  numerators  of 
the  several  fractions  as  directed  to  arrange  whole  numbers, 
when  whole  numbers  only  are  given,  and  place  each  denominator 
opposite  its  own  numerator. 

Note. — Before  stating  the  sum,  mixed  numbers,  if  any  are 
given,  must  be  reduced  to  improper  fractions. 

Ex  1 .  If  f  of  a  yard  cost  -fj  of  a  pound,  what  will  -^  of  a 
yard  cost  ? 

Statement,  =•='  ,V  Q- 
15.  14.  o 

The  J  is  inverted,  that  its  numerator  may  stand  below  the  line' 
as  the  same  term  would  stand  if  it  were  a  whole  number. 


PROPORTION  IN  FRACTIONS.  207 

Solution'  dhrl 

2 
Therefore,  ^  of  a  pound,  or  3  s.  4d.,  is  the  answer  required. 

2.  If  $  of  a  pound  of  sugar  cost  f  of  a  shilling,  what  will 
•A-  of  a  pound  cost  ? 

894 
Statement,  9'  1Q'  3>     Ans.  1  s.  0  d.  3J  qr. 

3.  A  person  who  owned  f  of  a  vessel  sold  f  of  his  share 
for  375  £.     What  was  the  value  of  the  whole  vessel,  at  the 
same  rate?     Ans.  1000  £. 

These  sums  may  all  be  solved  analytically,  if  preferred.    The 
following  is  the  solution  of  the  last  :    |  of  f  =  JJ,  and  375  £. 
-i-l5=25  «£.,  or  ^  of  the  whole  value  ;  therefore,  25  £.+40 
Ans. 


4.  If  f  of  a  ship  be  worth  3740  £.,  what  is  the  value  of 
the  whole  1     Ans.  9973  £.  6  s.  8  d. 

5.  If  1^  yards  cost  9  shillings,  what  is  the  value  of  16£ 
yards  ?     Ans.  5  £.  17  s. 

6.  What  is  the  value  of  |-  of  a  pound  of  lard,  if  -^  of  a 
pound  cost  £  of  a  shilling?     Ans.  2J&  pence. 

7.  A  person  who  owned  -|  of  a  lot  of  land,  sold  f  of  his 
share  for  $3024.    What  was  the  value  of  the  whole  lot,  at  the 
same  rate?     Ans.  $12096. 

8.  A  certain  vessel  is  valued  at  $1562.50.     What  is  the 
value  of  f  of  f  of  the  same  ?     Ans.  $500. 

9.  I  owned  f  of  a  ship,  and  sold  f  of  my  share  for  $780. 
What  was  the  value  of  the  whole,  at  the  same  rate  ?     Ans. 
$3120. 

10.  A  merchant  bought  5J  pieces  of  cloth,  each  containing 
24f  yards,  for  6|  shillings  per  yard.    How  many  dollars  did 
the  whole  cost,  in  New  York  currency?    Ans.  $111. 

11.  A  merchant  had  4f  cwt.  of  sugar,  at  6^  pence  per  lb., 
which  he  exchanged  for  tea,  at  8|  shillings  per  pound.     How 
many  pounds  of  tea  did  he  receive  ?     Ans.  29J. 

12.  How  many  pounds  sterling  will  150  yards  of  cloth  cost, 
at  1^  shilling  per  yard  ?     Ans.  9  £. 

13.  If  3^  times  3£  yards  cost  1^  times  Impounds  sterling, 
how  many  shillings  and  pence  will  ^  of  £  of  12£  yards  cost  ? 
.4ns.  7s.  6  d. 

14.  What  is  the  value  of  |  of  an  ounce  of  silver,  if  2  oz.  be 
valued  at  12|  shillings  ?     Ans.  4  s.  9  d. 


208  PROPORTION  IN  FRACTIONS. 

15.  What  quantity  of  shalloon,  that  is  £  of  a  yard  wide,  will 
be  sufficient  to  line  7£  yards  of  cloth,  l£  yards  wide  1     Ans. 
15  yards. 

16.  If  2^  yards,  li  yard  wide,  be  sufficient  to  make  a  coat, 
how  much  will  it  require  of  cloth  that  is  |  of  a  yard  wide  to 
make  the  same  kind  of  garment  1     Ans.  4  yd.  3^T  qr. 

17.  How  many  pieces  of  cloth,  at  $18f per  piece,  are  equal 
in  value  to  224f  pieces,  at  $12£  per  piece  ?     Ans.  $15Qi|. 

18.  A  merchant  exchanged  7$  cwt.  of  sugar,  at  7§  pence 
per  pound,  for  tea  at  9^  shillings  per  pound ;  how  many  pounds 
of  tea  did  he  receive  ?    Ans.  60f±  Ib. 

19.  If  8  men  can  perform  a  piece  of  work  in  6|  hours,  in 
what  time  will  20  men  do  the  same  ?     Ans.  2  hours,  40  min- 
utes. 

20.  How  many  yards  of  cloth,  f  of  a  yard  wide,  will  line  20 
yards,  f  of  a  yard  wide  ?     Ans.  12  yards. 

21.  How  many  pieces  of  cloth,  at  18^  shillings  per  yard, 
are  equal  in  value  to  350£  pieces,  at  12^  shillings  per  yard  ? 
Ans.  241|-J  pieces. 

22.  Lent  a  friend  $72f  for  8f  months  ;  what  sum  must  he 
lend  me  for  2£  years,  to  balance  the  favor?     Ans.  $21.233.-}- 

The  following  sums  properly  belong  to  Compound  Propor- 
tion. They  may  be  solved  either  by  canceling,  by  analysis,  or 
by  the  common  rule  of  Compound  Proportion. 

Ex.  23.  If  |  of  a  yard  of  cloth,  which  is  |-  of  a  yard  wide, 
cost  J  of  a  pound  sterling,  what  is  the  value  of  f  of  a  yard, 
that  is  l^yard  wide? 

Analysis :  |  X  f — fj,  the  fraction  of  a  square  yard  purchased, 
which  cost  f  of  a  pound  sterling.  Therefore,  f  •+  2 1  =3^5 ,  the 
value  of  Jj  part  of  a  square  yard,  and  -j-|j  X  32  =^3-,  the  price 
of  1  yard,  f  X  J=f|-,  the  quantity  of  which  the  price  is  requi- 
red. Therefore, T6^xffn:|||^=:|  of  1  £.  Ans.  =  13  s.  4d. 

The  same  canceled,  ^  ^  ]'  *'  g. 

O.    o.   4.   o.    / 

To  understand  why  the  fractions  |  and  £  are  inverted,  it 
must  be  remembered  that  a  fraction  is  divided  by  a  fraction,  by 
inverting  the  divisor  and  then  multiplying  numerators  and  de- 
nominators together.  (See  Sec.  7th,  Introduction  to  Fractions.) 

The  above  solved,  ^  I'  *'  *'  **      f  of  a  £.  =  13  s.  4  d.   Ans. 

S.   o..   4.   o.   " 

24.  If  9  men  spend  125«£.  in  27  days,  what  sum  will  25 
men  spend  in  40  days  ? 


CONJOINED  PROPORTION.  209 

Analysis  :  12 1  £.  =  *g  £.,  and  2-f  +-  9 =f|  £.  the  money  one 
man  spends  in  27  days  ;  and  ^|  £.  -^-27=-^^-,  the  money  spent 
by  one  man  daily.  Therefore,  -f/^  £.  x  25  =  Jff  £.  the  money 
25  men  spend  daily;  and  fff  £.  X40=-?|^oo  £  the  sum  of 
money  required,  which  reduced  gives  51  £.  8s.  9|-|- pence, 
Ans. 

™  •••*:-  ,.        25.  25.  40 

ihe  same  stated  tor  canceling,  —= —        . 

20 

Canceled,  ^-^pf^  and  25x25x20  =  12500;  and  27x9 
i=243;  and  12500^243  =  51  £.  8s.  9ffd.  Ans. 

25.  If  18  persons  consume  f^lb.  of  tea  in  one  month,  how 
much  will  8  persons  consume  in  six  months  ?     Ans.  4j  Ib. 

26.  If  the  tuition  of  2  boys  for  |  of  a  year  be  56|  £.,  how 
much  will  be  the  tuition  of  3  boys  for  5^  years  ?     Ans.  600  £. 

27.  If  90  cwt.  be  carried  30  miles  for  $29,  how  many  cwt. 
may  be  carried  45  miles  for  $5f  ?     Ans.  12  cwt. 

28.  If  10  persons  drink  15f  gallons  of  wine  in  one  week, 
how  much  will  16  persons  drink  in  43  weeks  ?     Ans.  1073^ 
gallons. 

29.  If  5  cwt.  be  carried  600|  miles  for  $12$,  how  far  may 
|  of  a  cwt.  be  carried  for  $30|  ?     Ans.  988 Jy  miles. 

QUESTIONS. — When  is  an  arithmetical  question  solved  analytically  1 
What  is  the  general  principle  by  which  sums  may  be  solved  analyti- 
cally 7  How  are  sums  in  Simple  or  Compound  Proportion  solved 
without  canceling  1  How  are  they  solved  by  canceling?  What  is 
the  note  1 


CONJOINED    PROPORTION. 

Conjoined  Proportion  consists  of  a  comparison  instituted 
between  a  series  of  terms  bearing  a  certain  relation  to  each 
other,  as  the  coins,  weights,  and  measures  of  different  coun- 
tries. 

The  principle  involved  in  this  rule  is  the  same  as  in  Single 
and  Compound  Proportion.  No  farther  explanation  is  there- 
fore needed, 

19* 


210  CONJOINED  PROPORTION. 

RULE. — Above  a  horizontal  line  near  the  left,  place  the  de- 
manding term ;  then  below  the  line,  place  the  term  of  the  same 
name  as  the  demand,  with  the  term  which  it  equals  in  value  di- 
rectly above  it.  Next,  seek  another  term  of  the  same  name  as 
the  one  last  placed,  and  set  it  also  below  the  line,  with  the  one  it 
equals  in  value  also  above  it.  Thus  proceed  to  arrange  the  terms, 
making  each  term  standing  below  the  line  of  the  same  name  as 
the  preceding  term  standing  above  it.  The  product  of  the  num- 
bers standing  above  the  line  divided  by  the  product  of  those 
standing  below  it,  will  give  the  required  number. 

The  numbers  may  of  course  be  canceled  as  far  as  practica- 
ble, before  multiplying  and  dividing. 

Ex.  1.  If  100  Ib.  English  make  90  Ib.  Flemish,  and  22  Ib. 
Flemish  make  28  Ib.  Bologna,  how  many  pounds  English  are 
equal  to  56  Ib.  Bologna  ? 

The  demand  obviously  lies  on  the  56  Ib.  Bologna;  therefore, 
56.  22.  100 
88.^5' 
2 

Canceled,  **'  ^'  ^     2x22x  10=440+-9=:48f  Ib.   Eng- 
lish, Ans. 

2.  If  40  Ib.  at  New  York  make  48  Ib.  at  Antwerp,  and  30 
Ib.  at  Antwerp  make  36  at  Leghorn,  how  many  pounds  at  New 
York  are  equal  to  144  at  Leghorn  ? 

144.  30.  40 
Statement, ^-^. 

4     5     20 

Canceled,  -     '  ^  "4&  ^  ^   and  5x20  =  100  Ib.   New  York, 
Ans. 

3.  If  70  braces  at  Venice  make  84  braces  at  Leghorn,  and 
12   at  Leghorn  make  7  American  yards,  how  many  braces   at 
Venice  are  equal  to  96  American  yards  1     Ans.   1371. 

4.  If  24  Ib.  at  New  London  make  20  Ib.  at  Amsterdam,  and 
50  Ib.  at  Amsterdam  make  60  Ib.  at  Paris,  how  many  Ib.  at  Paris 
are  equal  to  40  Ib.  New  London  ?     Ans.  40  Ib. 

5.  If  50  Ib.  at  New  York  make  45  Ib.  at  Amsterdam,  and  80  Ib. 
at  Amsterdam,   103  Ib.  at  Dantzic,  how  many  Ib.   at  Dantzic 
are  equal  to  240  Ib.  New  York  ?     Ans.  278^. 

6.  If  24  braces  at  Leghorn  be  equal  to  15  vares  at  Lisbon, 
and  45  vares  at  Lisbon  be  equal  to  90  braces  at  Lucca,  how 


DISCOUNT.  -  211 

many  braces   at  Lucca  are  equal  to  120  braces  at  Leghorn  ? 
Ans.  150  braces. 

QUESTIONS. — In  what  does  Conjoined  Proportion  consist  7  How 
does  the  principle  involved,  compare  with  Simple  and  Compound 
Proportion  1  What  is  the  rule  for  Conjoined  Proportion  1 


DISCOUNT. 

Discount  is  an  allowance  made  for  the  payment  of  money 
before  it  becomes  due. 

The  present  worth  of  any  sum  of  money,  payable  at  some 
future  time  without  interest,  is  that  sum  which,  if  put  at  inte- 
rest, would  in  the  given  time  and  rate  per  cent,  amount  to  the 
whole  debt. 

Discount  is  not,  therefore,  a  deduction  of  the  given  per  cent, 
from  a  hundred  cents  or  a  hundred  dollars.  If  I  have  a  claim 
upon  an  individual  for  $100,  payable  a  year  hence ;  and  propose 
to  allow  him  6  per  cent,  discount  for  present  payment,  I  must 
receive  more  than  $100  —  $6  =  $94  ;  since  $94  put  on  inte- 
rest at  6  per  cent,  will  not  amount  to  $100  in  the  given  time. 
The  interest  on  $94  one  year  at  6  per  cent,  is  $5.64  ;  and 
$94  + $5. 64=: $99.64,  which  is  36  cents  less  than  the  re- 
quired sum,  or  $100.  If,  however,  a  person  owe  me  $106,  pay- 
able in  one  year  without  interest,  and  I  propose  to  allow  him 
the  same  discount  for  immediate  payment,  he  must  obviously 
pay  me  $100,  since  $100  in  one  year  at  six  per  cent,  will 
amount  to  precisely  $106. 

Hence,  we  learn  that  the  ratio  which  any  sum  due  a  year 
hence  without  interest  bears  to  its  present  worth,  is  as  1 06  to 
100 ;  or,  what  is  the  same  thing,  as  $1.06  to  $1.00,  whenever 
the  discount  is  at  6  per  cent.  If  the  rate  per  cent,  be  any 
other  than  6,  or  the  time  more  or  less  than  one  year,  the  ratio 
varies  accordingly.  Therefore,  as  the  amount  of  $  1  for  the 
given  time  and  rate  per  cent,  is  to  $l,so  is  the  given  sum  to  its 
present  worth. 

Ex.  1 .  What  is  the  present  worth  of  $450,  due  2  years 
hence,  6  per  cent,  discount  being  allowed  ? 


212 


DISCOUNT. 


The  interest  of  $1  for  2  years  at  6  per  cent,  is  12  cents, 
and  consequently  the  amount  of  $1,  for  the  same  time,  is  $1.12. 
Therefore,  1.12:1  ::  450:  the  required  sum.  And,  since 
nothing  is  effected  by  multiplying  by  1 ,  the  required  sum  is 
obtained  by  dividing  $450  by  $1.12.  Hence,  $150.00  — 
$1.12  =  $401.785.+  Ans. 

From  the  above  we  derive  the  following  rule  : 

RULE. — Divide  the  sum  on  which  the  discount  is  to  be  made, 
by  the  amount  of  one  dollar  for  the  given  time  and  rate  per 
cent. 

2.  What  is  the  present  worth  of  $700,  due  3  years  hence, 
at  5  per  cent,  discount  ? 

The  amount  of  $1  for  3  years  at  5  per  cent.,  is  $1.15. 
Therefore,  $700.00 -=-'l.l  5= $608.695. -f  Ans. 

3.  Sold  goods  to  the  amount  of  $1200,  on  6  months'  credit. 
What  is  the  present  worth,  allowing  8  per  cent,  discount  ? 
Ans.  $1153.846.+ 

4.  What  is  the  present  value  of  a  legacy  of  $2000,  due  2 
years  hence,  discounting  at  5  per  cent,  per  annum  ?     Ans. 
$1818.18.+ 

5.  What  is  the  difference  between  the  interest  and  discount 
on  $600  for  12  years,  at  5  per  cent.?     Ans.  Interest  $360; 
discount,  $225  ;  difference,  $135. 

Note. — To  obtain  the  discount,  subtract  the  present  value 
from  the  sum  due. 

6.  What  is  the  discount  on  $300  for  60  days,  at  6  per  cent, 
per  annum?     Ans.  $2.97. 

7.  What  is  the  present  value  of  $750,  due  3|  years  hence, 
discounting  at  4  per  cent,  per  annum?     Ans.  $657. 894. -f- 

8.  What  is  the  discount  on  $500  for  2  years,  at  9  per  cent, 
per  annum?     Ans.  $76.272.+ 

9.  What  is  the  present  value  of  350  £.,  due  4  years  hence, 
discounting  at  4  per  cent,  per  annum  ?     Ans.  301  £.  14s.  5  d. 

3inr  9r- 

10.  What  is  the  present  worth  of  672  £.,  due  2  years  hence, 

discounting  at  the  rate  of  6  per  cent,  per  annum  ?  Ans.  600  £. 

1 1 .  Bought  goods  to  the  amount  of  $820,  on  6  months'  credit. 
What  ought  I  to  have  paid,  if  I  had  advanced  the  money  on 


PROFIT  AND  LOSS.  213 

the  receipt  of  the  goods,  and  had  been  allowed  4  per  cent,  dis- 
count?    Ans.  $803.92.+ 

12.  Sold  goods  to  the  amount  of  81200,  one  half  of  which 
is  to  be  paid  in  6  months,  and  the   other  half  in  8  months. 
What  is  the  discount  for  the  present  payment  of  the  whole, 
discounting  at  6  per  cent,  per  annum  ?     Ans.  $40.553. 

13.  A  person  having  a  legacy  of  $1450  left  him,  payable  in 
6  years,  requests  present  payment,  and  proposes  to  allow  6 
per  cent,  discount.     What   must  he  receive?     Ans.  $1066.- 
176.+ 

14.  What  is  the  discount  of  $458  for  8  months,  discounting 
at  8 per  cent,  per  annum?     Ans.  $23.188.+ 

Q.UESTIONS. — What  is  discount  1  What  is  the  present  worth  of  any 
sum  of  money,  payable  at  some  future  time,  without  interest  1  Is  dis- 
count a  deduction  of  a  given  per  cent,  from  a  hundred  cents,  or  a 
hundred  dollars'?  Why  1  What  numbers  express  the  ratio  which 
any  sum  due  a  year  hence,  at  6  per  cent,  bears  to  its  present  worth  7 
What  is  the  rule  for  discount  1  How  is  the  discount  obtained  1  How 
is  discount  proved  1  Ans.  Cast  the  interest  on  the  present  worth,  for 
the  time  and  rate  per  cent,  of  discount,  and  add  it  to  the  present  worth. 


PROFIT    AND   LOSS. 

Profit  and  Loss  is  the  rule  by  which  merchants  and  others 
engaged  in  trade,  determine  how  much  is  gained  or  lost  by  any 
transaction.  It  also  enables  them  so  to  regulate  the  price  of 
their  goods  as  to  gain  or  loose  a  certain  per  cent,  on  the  first 
cost. 

1st.  To  FIND  HOW  MUCH  IS  GAINED  OR  LOST  ON  A  QUANTITY 
OF  GOODS  SOLD  AT  RETAIL,  THE  PURCHASE  PRICE  OF  THE 
WHOLE  QUANTITY  BEING  GIVEN. 

RULE. — Find  the  value  of  the  whole  quantity  at  the  retail 
price;  then,  if  there  be  a  gain,  subtract  the  purchase  price  from 
the  same,  and  the  remainder  will  be  the  sum  gained ;  but  if 
there  be  a  loss,  subtract  the  amount  received  from  the  purchase 
price,  and  the  remainder  will  be  the  sum  lost. 


214  PROFIT  AND  LOSS. 

Ex.  1.  Bought  40  yards  of  cloth  for  $160,  and  sold  the 
same  for  $5.20  per  yard  ?  How  much  did  I  gain  ?  $5.20  X  40 
=  $208.00;  and  $208.00  — $160  =  $48.00,  Arts. 

2.  Bought  a   hogshead  of  melasses  for  $25,  and  sold  the 
same  for  8  cents  a  pint.    How  much  did  I  gain  ?  Ans.  $15.32. 

3.  Bought  12  cwt.  of  sugar  at  8  d.  per  pound,  and  sold  it  at 
3  £.  New  York  currency,  per  cwt.     Did  I  gain  or  lose,  and 
how  much?     Ans.  Lost  $22. 

4.  Purchased  2  hogsheads  of  wine  for  $94.50,  and  retailed 
the  same  at  2  s.  New  York  currency,  per  quart.     How  much 
did  I  gain?     Ans.  $31,50. 

5.  Paid  $57  for  456  yards  of  cloth,  and  sold  the  same  at 
the  rate  of  4  s.  6  d.  New  York  currency,  for  3  yards.     What 
did  I  gain?     Ans.  $28.50. 

6.  Bought  12  rolls  of  ribbon,  each  containing  50  yards,  for 
$18.75,  and  sold  the   same  at  6d.  New  York  currency,  per 
yard.     How .  much  did  I  gain  by  the  operation  ?     Ans.  $  1 8.75 . 

7.  Bought  44  Ib.  of  tea  for  $16.50,  and  sold  it  for  3  s.  6  d. 
New   England  currency,  per  pound.     What  did  I  gain  ?  Ans. 
$9.166.+ 

8.  What  do  I  gain  on  15  cwt.  of  rice,  which  cost  me  $50, 
by  retailing  the  same  at  4  d.  New  York  currency;  per  pound  ? 
Ans.  $20. 

2d.    To  FIND  WHAT  IS  GAINED  OR  LOST  PER  CENT 

RULE. — Find  the  whole  gain  or  loss  by  the  preceding  rule, 
and,  having  multiplied  it  by  one  hundred,  divide  the  product  by 
the  first  cost.  Or  say,  as  the  first  cost  is  to  the  whole  gain  or 
loss,  so  is  $100  or  ]00j£.  to  the  gain  per  cent. 

Ex.  1.  If  I  buy  broadcloth  at  $5.50  per  yard,  and  sell  the 
same  for  $6.00  per  yard,  what  do  I  gain  per  cent.  ? 

$6.00 — $5.50  —  $0.50,  the  gain  on  $5.50.  Therefore,  .50 
X  100=50.00,  and  50.00-^- $5.50  =  9.09+,  the  gain  on  $100. 
Or,  $5.50  :  .50  :  :  100  :  the  gain  on  $100. 

Or,  the  operation  may  be  canceled,  by  -p\a,cmgf,rst  above  the 
horizontal  line,  the  whole  gain  or  loss  found  by  subtraction,  and 
$100  or  100  £.  at  the  right  of  this,  on  the  same  side,  and  the 
whole  cost  below  the  same.  Cancel,  (Sf-c. 

The  above  sum  thus  stated,  '5|4ir»  and  100 -f- 11  =$9.09  +  , 

a  .50. 
11 
as  before. 


PROFIT  AND  LOSS.  215 

2.  What  do  I  gain  per  cent,  if  I  buy  wheat  at  12  s.  a  bushel, 
and  sell  the  same  for  15  s.  a  bushel  ?     Ans.  25  per  cent. 

3.  Purchased  pepper  for  8d.  per  pound,  and  sold  the  same 
for  9  d.  per  pound.     What  per  cent,  did  I  gain  ?    Ans.  12%. 

4.  Bought  650  Ib.  of  sugar  for  10  cents  a  lb.,  and  sold  the 
same  for   12  cents  per  lb.     What  was  my  whole  gain,  and 
what  my  gain  per  cent.  ?     Ans.  Whole  gain   $13.00  ;  gain 
per  cent.  $20. 

5.  Bought  goods  to  the  amount  of  $325,  and  sold  the  same 
for  $370.     What  was  the  per  cent,  gained  ?     Ans.  $  1 3.846.  + 

6.  If  I  lose  $2  on  $25,  at  what  rate  per  cent,  do  I  lose  ? 
Ans.  8  per  cent. 

7.  Purchased  a  hogshead  of  wine   for   $50,  and  sold  the 
same  for  $75,  on  six  months'  credit.     What  was  my  gain  per 
cent.,  allowing  4  per  cent,  discount  for  the  6  months'  credit  ? 
Ans.  $44.23. 

8.  Bought  6  cwt.  of  cheese  for  $48,  but  it  being  damaged,  I 
am  willing  to  sell  it  for  the  same  on  a  year's  credit.     What  is 
my  loss  per  cent,  discounting  at  6  per  cent,  per  annum  1  Ans. 
$5.66.+ 

3d.    To  FIND    HOW    A    COMMODITY   MUST    BE    SOLD    TO   GAIN  OR 
LOSE  A  CERTAIN  PER  CENT.  ON  THE  WHOLE  COST. 

RULE. — If  the  purchase  price  of  the  quantity  for  which  the 
retail  price  is  required,  be  not  given,  it  must  first  be  found,  and 
then  multiplied  by  $100  increased  by  the  per  cent,  to  be  gained, 
or  diminished  by  the  per  cent,  to  be  lost.  The  last  product  di- 
vided by  100,  or,  what  is  the  same  thing,  with  two  figures  cut  of 
from  the  right,  will  be  the  answer  required.  Or  the  operation 
may  be  reduced  to  the  following  statement :  As  $100  is  to  $100 
increased  by  the  per  cent,  to  be  gained,  or  diminished  by  the  per 
cent,  to  be  lost,  so  is  the  purchase  price  of  the  commodity  to  the 
retail  price. 

Ex.  1.  Bought  300  yards  of  cloth  for  $550;  how  must  I  sell 
the  same  per  yard,  to  gain  25  per  cent.  ? 

$550-=-300=$1.83J,  the  price  of  one  yard;  and  $1.83$  x 
125  =  $229.16;  and  $229.16-:- 100  =  $2.29,+  Ans.  Or  100  : 
125  :  :  1.83$  :  Ans.  or  $2.29.+ 

Or  the  operation  may  be  canceled  by  the  following  rule  : 

RULE. — Write  the  given  price  above  a  horizontal  line,  and 
the  quantity  which  cost  that  price  directly  below  it.  Then  place 


216  PROFIT  AND  LOSS. 

$100  increased  by  the  per  cent,  to  be  gained  or  diminished  by 
the  per  cent,  to  be  lost,  above  the  same  line,  and  100  below,  and 
proceed  to  cancel,  <fyc. 

The  above  sum  stated,  |^^. 

5 
Canceled,         -,  and  55-6X4  =  $2.29.+ 


2.  How  must  I  retail  melasses  by  the  gallon,  for  which  I 
paid  $30  per  hogshead,  to  gain  12|  per  cent.  ?  Ans.  $0.535.4- 

3.  Purchased  100  gallons  of  wine  for  $130,  but  by  accident 
15  gallons  leaked  out.    How  must  I  sell  the  remainder  per  gal- 
ion,  to  gain  15  per  cent.  1     Ans.  $  1.758.  -+- 

4.  Paid  $1.10  per  gallonfor  melasses  ;  how  must  I  sell  the 
same  per  quart,  to  gain  30  per  cent.  ?     Ans.  35|  cents. 

5.  Received  from  Lisbon  180  casks  of  raisins,  which  cost 
me  $2.13  per  cask.    How  shall  I  sell  them  per  cask,  to  gain  25 
per  cent.  ?     Ans.  $2.66. 

2.13.  125 
Statement,  --  —  . 

6.  Bought  2  cwt.   of  pepper  at  1  s.   New  York  currency, 
per  pound,  but  it  being  damaged,  I  am  willing  to  lose  10  per 
cent.  ;  how  must  I  sell  it  by  the  Ib.  1     Ans.  1  1  cents  2^  mills. 

7.  Bought  one  ton  of  steel  for  $184  ;  how  must  I  sell  the 
same  per  pound,  to  gain  5  per  cent.  1     Ans.  8  cents  6  mills. 

8.  A  merchant  bought  160  yards  of  cloth   for  $240;  how 
must  he  sell  the  same  per  yard  to  gain  12  per  cent.  ?     Ans. 
$1.68. 

9.  Bought  8  pieces  of  cloth,   each  containing  15  yards,  at 
3  s.    New   England  currency,  per  yard  ;   how  must  I  retail 
the  same,  to  gain  8  per  cent.,  and  how  much  must  I  receive 
for  the  whole  ?      Ans.   54  cents  retail  price  ;    whole  value, 
$64.80. 

10.  If  I  buy  6  cwt.  of  sugar  at  10  d.  New  York  currency, 
per  pound,  and  am  allowed  4  per  cent,  discount  for  ready 
money,  and  sell  the  same  so  as  to  gain  15  per  cent,  on  the 
money  advanced,  how  much   money  do   I  receive  ?      Ans. 
$77.28. 

11.  Bought  12  chests  of  tea,  each  weighing  56  pounds,  at 
4s.   6  d.  New   England  currency,  per  Ib.     For  ready  money 
was  allowed  2  per  cent,  discount  ;  how  much  must  I  receive 


BARTER.  217 

for  the  whole  to  realize  a  profit  of  10  per  cent,  on  the  money 
paid  out?     Ans.  $543.312. 

12.  56.  54.  98.  110 
Statement,  -     12.   6.  m  1(X). 

12.  Bought  700  yards  of  ribbon,  at  6  d.  New  York  currency, 
per  yard ;   how  must  I  sell  the  same  to  gain  12  \  per  cent,  and 
what  shall  I  receive  for  the  whole  quantity  ?     Ans.  6|  d.  per 
yard,  nearly;  and  for  the  whole,  $49.218.+ 

13.  How  must  cloth  which  costs  13  s.  4  d.  be  sold  to  gain  20 
per  cent.?     Ans.  At  16  s.  per  yard. 

CtuEsrioNS. — What  is  profit  and  loss  1  What  is  Case  1st  7  What 
is  the  rule?  What  is  Case  2d?  What  is  the  rule?  How  is  the 
statement  made  for  canceling  1  What  is  Case  3d  1  What  is  the  rule  ? 
What  is  the  rule  for  canceling  1 


BARTER. 

Barter  is  a  rule  by  which  those  engaged  in  trade  exchange 
commodities  so  that  neither  party  suffer  loss. 

Take  the  following  illustration : 

Ex.  1.  How  much  sugar  at  lOd.  per  Ib.  must  be  given  in 
exchange  for  12  cwt.  of  butter  at  15  d.  per  pound? 

It  is  obvious  that  the  12  cwt.  of  butter  bears  the  same  ratio 
to  the  number  of  cwt.  of  sugar  required,  as  10  d.  bears  to  15  d. 

Therefore,  as  10 : 15 :  :12 :  the  answer,  viz.  18  cwt.  Or,  |jp^  is 
the  statement  for  canceling  : 
6      3 

^~f,  6x3  =  18,  Ans. 
S 

The  scholar  will  perceive  that  the  principle  here  involved 
is  the  same  as  in  the  rule  of  proportion,  which  has  already 
been  fully  explained. 

2.  How  much  tea  at  8  s.  per  Ib.  must  be  given  for  2  cwt.  of 
chocolate  at  4  s.  6  d.  per  Ib.  ?     Ans.  I  cwt.  14  Ib.   ^ 
19 


218  BARTER. 

Statement,  8  s. :  4  s.  6  d.  : :  2  cwt. :  Ans. 

-r.  v  2.54 

For  canceling,  — TO~Q- 

In  this  statement,  the  price  of  the  chocolate  is  reduced  to 
pence.     The  12  below  the  line  reduces  the  pence  to  shillings. 

3.  How  much  rice   at  32  s.  per   cwt.  must  be  given  for  4 
cwt.  2  qr.  of  raisins   at  6  d.   per  pound  ?     Ans.  7  cwt.  3  qr. 
14  Ib. 

4.  B.  has  90  yards  of  linen,  worth  4  s.  6  d.  per  yard,  which 
he  wishes  to  exchange  with  C.  for  muslin  at  2  s.  per  yard  ; 
howmany  yards  of  muslin  must  B.  receive  ?    Ans.  202|  yards. 

5.  Bought  75  yards  of  broadcloth  at  $3  per  yard,  and  paid 
for  it  in  muslin  at  1  s.  3  d.  New  York  currency,  per  yard  ;  how 
many  yards  of  muslin  did  it  take?     Ans.  1440  yards. 

6.  Bought  7  tuns  of  wine  at  10  d.  a  pint,   and  paid  for  it 
with  melasses  at  3  s.  per  gallon  ;  how  many  hogsheads  of  me- 
lasses  did  it  take?     Ans.  62|  hhd. 

7.  Sold  5  cwt.  1  qr.  of  sugar  at  $8.50  per  cwt.  and  received 
in  pay  24  yards  of  cloth ;  what  was  the  value  of  the  cloth  per 
yard?     Ans.  $1.859. 

8.  How  many  gallons  of  melasses  at  8  cents  a  quart  must 
be  given  for  24  cwt.  of  rice  at  $8  per  cwt.     Ans.  600  gallons. 

9.  I  have  nuts  worth  $4  per  bushel,  and  B.  has  sugar  worth 
10  cents  per  pound ;  now  if  I  charge  him  $4.50  per  bushel 
for  my  nuts,  what  ought  he  to  charge  me  for  his  sugar  ?     Ans. 
11^  cents. 

10.  What  quantity  of  butter  worth  12|  cents  a  pound  must  be 
given  in  exchange  for  12  Ib.  of  indigo  worth  $2.25  per  pound  ? 
Ans.  216  Ib. 

1 1 .  Sold  5  cwt.  of  cheese  at  8  d.  per  pound,  and  received  in 
pay  muslin  worth  Is.   6  d.  per  yard ;  how  many  yards  did  I 
receive  ?     Ans.  248|  yds. 

12.  Bought  90  bushels  of  wheat  at  15  s.  and  gave  in  pay  115 
yards  of  Irish  linen,  worth  6  s.  per  yard,  and  the  remainder  in 
cash;  how  many  dollars  did  I  pay,  the  currency. being  New 
York  ?     Ans.  $82.50 

13.  Received  of  B.  a  quantity  of  broadcloth  which  he  had 
sold  at  $4.50  per  yard,  but  charged  me  $5.00;  in  pay  I  let 
him  have  a  quantity  of  wheat  which  I  had  retailed  at  $1.50 
per  bushel ;  what  ought  I  to  have  charged  him,  to  meet  the  ad- 
vance he  made  on  his  broadcloth?     Ans.  $l.66f. 

14.  Bought  12  gross  of  pen-knives  at  9  dollars  per  gross,  and 


PARTNERSHIP.  219 

paid  for  them  in  cloth  at  3s.  6  d.  New  York  currency,  per 
yard  ;  how  many  yards  will  be  required  to  pay  for  the  knives  ? 
Ans.  246-|  yards. 

15.  Sold  48  cwt.  of  hops  at  4  d.  per  pound,  and  received  pay 
in  prunes  at  6  d.  per  pound  ;  how  many  cwt.  did  I  receive  ? 
Ans.  32  cwt. 

16.  Received  460  yards  of  linen  worth  6s.  6  d.  New  York 
currency,  per  yard,  in  exchange  for  84  yd.  of  broadcloth ; 
what  was  the  value  of  the  broadcloth  per  yard  ?  Ans.  $4.449.  + 

17.  Exchanged  320  Ib.  of  chocolate  valued  at  4s.  6d.  New 
York  currency,  per  pound,  for  a  quantity  of  cotton  at  8  d.  per 
Ib. ;  but  there  not  being  cotton  enough  to  pay  me,  I  received 
the  balance,  viz.  50  dollars,  in  cash ;  how  many  Ib.  of  cotton 
did  I  receive  ?     Ans.  1560  Ib. 

18.  How  many  cwt.  of  sugar,  valued  at  10  d.  per  Ib.  New 
England  currency,  must  be  given  in  exchange  for  24  cwt.  of 
pepper  at  15  dollars  per  cwt.  ?     Ans.  23^  cwt. 

19.  What  deduction  ought  to  be  made  on  wheat,  which  has 
been  valued  at  9  s.   per  bushel,  in  exchange  for  cloth,  when 
cloth  valued  at  14  s.  per  yard  is  sold  for  12  s.  per  yard  ?    Ans. 
Is.  3^d.  per  bushel. 

20.  How  many  pounds  of  cinnamon  at  10  s.  per  Ib.  must  be 
given  in  exchange  for  5  cwt.  of  saleratus  at  9  pe^ce  per  Ib.  ? 
Ans.  42  Ib. 

QUESTIONS. — What  is  Barter  1    What  is  the  principle  involved  in 
this  rule  ? 


PARTNERSHIP. 

Partnership  is  the  rule  by  which  individuals  trading  with  a 
joint  stock,  determine  their  several  shares  of  the  profit  or  loss 
realized. 

The  individuals  thus  trading  are  called  Stockholders.  The  sum 
of  money  advanced  by  the  company  forms  a  Capital  Stock ;  and 
the  money  from  time  to  time  received  on  the  several  shares  is 
called  Dividend.  When  the  whole  stock  owned  by  the  com- 


220  PARTNERSHIP. 

pany  is  employed  for  the  same  period  of  time,  the  gain  or  loss 
which  results  from  the  adventure  is  divided  among  the  indi- 
viduals composing  the  company,  in  proportion  to  their  several 
shares  of  the  stock. 

We  have  then  the  following  rule  : 

RULE.  —  As  the  whole  stock  is  to  each  man's  share  of  the 
stock,  so  is  the  whole  gain  or  loss  to  each  man's  share  of  the 
gain  or  loss. 

Ex.  1.  Three  persons  traded  in  company.  A's  stock  was 
$1200  ;  B's  stock  was  $4800,  and  C's,  $2000  :  they  gained 
$800;  what  was  each  man's  share  of  the  gain  ? 

$1200+$4800+$2000=8000,  amount  of  the  stock  ;  there- 
fore, 8000  :  1200  :  :  800:  A's  share,  viz.  $120.  Then  to  ob- 
tain B's  share,  8000  :  4800  :  :  800  :  B's  share,  or  $480.  And 
lastly,  for  C's  share,  8000  :  :  2000  :  :  800  :  C's  share,  or  $200. 

To  state  sums  for  canceling  : 

RULE.  —  Write  the  whole  gain  or  loss  above  the  horizontal 
line,  with  the  stock  of  one  of  the  company  standing  at  the  right 
of  it  ;  then  place  the  whole  stock  below  the  line,  and  proceed  to 
cancel,  <fyc. 

fiflO    1  200 

The  above  sum  thus  stated:  -      —=$120,    A's  share. 

oUUU 


s$480,  B's  share  ;  and  ???~=  $200,  C's  share. 


Note.  —  The  operations  in  this  rule  are  proved  by  ad- 
ding together  the  several  shares.  Their  sum  must  equal  the 
whole  gain.  $120  +  $480  +  $200=$800. 


2.  Three  men,  A.,  B.,  and  C.,  traded  in  company.    A.  put  in 
$560;  B.,  $700  ;  and  C.,  $640.  At  the  expiration  of  their  part- 
nership, they  found  that  they  had  gained  only  $420  ;  what  was 
each  man's  share?  Ans.  A's,  $123.789;  B's,  $154.737;  and 
C's,  $141.473. 

3.  Three  merchants  trading  in  company  suffered  a  loss  of 
$600;  their  several  stocks  were  $800,  $1000,  and  $1200; 
what  was  each  man's  loss  ?     Ans.  $160,  $200,  and  $240. 

4.  A.,  B.,  and  C.,  freighted  a  ship.    A.  put  on  board  36  tons  ; 


PARTNERSHIP.  221 

B.,  40  tons  ;  and  C.,  60  tons.  In  a  storm,  30  tons  were 
thrown  overboard.  What  Avas  each  man's  share  of  the  loss  ? 
Ans.  A's,  ?if  tons  ;  B's,  8if  tons  ;  and  C's,  13r4y  tons. 

5.  A  bankrupt,  who  has  property  valued  at  $2500,  owes  A. 
$900  ;  B.,  1000  ;  C.,  $800,  and  D.,  $500.     How  much  will 
each  man  receive,  if  the  whole  property  be  given  up  to  them  ? 
Arts.  A.,  $703.125;  B.,  $781.25;  C.,  $625  ;  and  D.,  $390.625. 

6.  Three  individuals  unite  in  purchasing  600  acres  of  land. 
A.  pays  $3000  ;  B.,  $5000  ;  and  C,  $7000.     What  are  their 
several  proportions  of  the  whole  ?     Ans.  A.  has  120  acres  ;  B., 
200;  andC.,  280. 

7.  Three  individuals,  L.,  M.,  and  N.,  freighted  a  ship  with 
324  tons  burden.      L.   put  on  board  72  tons;  M.,    108  tons; 
and  N.,  144  tons.     They  cleared  $1200  by  the  adventure. 
What  was  each  man's   share  of  the  gain  ?     Ans.  L's,  $266.- 
666+  ;   M's,  $400  ;  and  N's,  $533.333.+ 

8.  Three  persons  agreed  to  pay  $  1 50  for  the  use  of  a  pas- 
ture, and  that  each  should  pay  in  proportion  to  the  number  of 
creatures  he  put  in      What  would  be  their  several  shares,  pro- 
vided the  first  put  in  100  oxen  ;  the  second,  150  oxen,  and  the 
third,  200  oxen?     Ans.  $33.333+  ;  $50  ;  and  $66.666.+ 

9.  A  person  at  his  death  owes  A.  70  £. ;  E.,  400  £. ;  and  H., 
140JE. ;    and  his  property  is  valued  at  only  400  £.       How 
much  ought  each  to  receive?     Ans.  A.,  45  £.  18^rs. ;  E., 
262  £.  5ff  s.  ;  and  H.,  91  £.  16^-s. 

10.  Four  individuals  purchased  a  horse  for  $96.     A.  paid 
$20;    B.,   $30;    C.,  25;    D.,   $21.     They  sold  the  horse 
for  $150.     What  was  each  man's   share  of  the  gain?     Ans. 
A's,  $11.25  ;  B's,  $16.875  ;  C's,  $14.06£;  and  D's,  $12.8l£. 

11.  A  man,  whose  property  was  worth  only  $500,  owed 
G.,  $150  ;  H.,  $180  ;  and  M.,  $300.     He  therefore  concluded 
to  give  his  property  up  to  his  creditors.     What  did  each  re- 
ceive ?    Ans.  G.,  $119.047+  ;  H.,  $142.857  ;  M.,  $238.095. 

12.  L.  has  a  farm  of  90  acres ;  M.  has  one  of  120  acres; 
and  N.  one  of  150  acres.     Their  farms  all  joining,  they  agree 
to  unite  them  for  the  purpose  of  renting.     This  they  do,  and 
receive  for  the  whole  the  annual  rent  of  $820  ;  what  is  each 
man's  share  of  the  rent  ?     Ans.  L's,  $205  ;  M's,  $273.333+  ; 
N's,  $341.666.+ 

Whenever  the  different  shares  are  not   continued  through 
the  same  period  of  time,  it  is  obvious  that  each  man's  share  of 
the  profit  and  loss  will  depend  not  entirely  on  his  portion  of  the 
19* 


222  PARTNERSHIP. 

whole  stock,  but  also  on  the  time  during  which  his   stock  was 
invested.    Take  the  following  sum  as  an  illustration  : 

13.  Three  merchants  traded  in  company.     A.  put  in  $120 
for  10  months  ;  B.,  $100  for  18  months  ;  and   C.,    $150  for  5 
months.     They  gained  $100.     What  was  each  man's  share  ? 
A's,  $120  for  10  months  =  $120  X  10  =  $1200  for  one  month  ; 
B's,  $100  for  18  months  =  $100x1 8  =  $1800  for  one  month; 
and  C's,  $150  for  5  months  =  $150x5  =  $750  for  one  month. 
Therefore,  $1200+$1800  +  $750  =  $3750,  and  3750:  1200:: 
100  :  A's  share,  or   $32.      3750  :  1800  : :  100  :  B's  share,  or 
$48  ;   and  3750  :  750  :  :  100  :  C's,  share,  or  $20. 

Hence,  when  the  time  of  investment  of  the  several  shares 
differs : 

RULE. — Multiply  each  man's  stock  by  the  time  during  which 
he  continued  in  trade,  and  use  the  several  products  as  directed 
in  the  preceding  rule  to  use  the  several  shares  of  the  whole  stock. 

14.  A.,  B.,  and  C.,  hold  a  pasture  in  common,  for  which 
they  pay  30  «£.  per  annum.     In  this  pasture,  A.  has  40  oxen 
for  75  days  ;  B.,  45  oxen  for  50  days  ;  and  C.,  50  oxen  for  90 
days.     What  part  of  the  30  £.  ought  each  to  pay  ?     Ans.  A., 
9£.  4s.   7^-d.  ;    B.,  6  £.  18  s.  5T73  d.  ;  andC.,   13  j£.  16s. 

aid. 

15.  Three  men  enter  into  partnership  on  the  following  terms  : 
A.   invests  $1500  for  5  months  ;  B.,  $1800  for  6  months  ;  and 
C.,  $2000  for  8  months.     During  the   continuance  of  their 
partnership  they  sustain  a  loss  of  $1000.     What  is  each  man's 
share  of  the  loss  ?     Ans,  A's  loss  is  $218.658  ;  B's,  $314.- 
868  ;  and  C's,   $466.472.      s 

16.  E.  and  S.  enter  into  partnership  for  one  year.     E.  at 
first  advances  $480,  and  B.  puts  in  his  share  3  months  after. 
How  much  must  he  advance  to  be  entitled  to  an  equal  share  of 
the  gain  at  the  expiration  of  one  year  ?     Ans.  $640. 

17.  Two  merchants  trading  in  company  gain  $200.     A's 
stock  was  $220  for  6  months,  and  B's,   $380  for  9  months. 
How   ought  they  to  share  the  gain  1     Ans.   A's  portion  is 
$55.696;  B's,  $144.304. 

18.  Two  men  commenced  trading  in  company  on  Jan.  1st, 
1825.   A.  advanced  $1000  at  the  time  specified;  but  B.,  finding  it 
inconvenient,  did  not  advance  his  share  till  the  first  of  May  fol- 
lowing.    At  the  end  of  the  year  they  shared  the  profits  equally. 
What  capital  did  B.  advance  I     Ans.  $1500. 


COMMERCIAL  EXCHANGE. 


223 


19.  A.  and  B.  traded  in  company.    A.  put  in  $1200  ;  B.  ad- 
vanced his  share  3  months  after.     What  sum  was  it  necessary 
for  him  to  advance  so  as  to  be  entitled  to  one  half  of  the  profit 
at  the  expiration  of  one  year  1     Ans.  $1600. 

20.  L.,  M.,  and  N.,  entered  into  partnership  ;  L.  advanced 
$300  for  3  years ;  6  months   after,  M.  put  in  $450 ;  and  6 
months  after  M.  put  in  his  share,  N.  put  in  $520.     At  the  ex- 
piration of  3  years,  they  found  they  had  cleared  $900.     What 
was   each  man's   share  1     Ans.   L's,   $264.274 ;  M's  share, 
$330.342  ;  and  N's,  $305.383. 

QUESTIONS. — What  is  Partnership  ?  What  are  the  individuals  tra- 
ding in  company  called  1  What  does  the  money  advanced  by  the 
company  form  1  What  is  the  money  they  receive  on  their  several 
shares  called  1  What  is  the  rule  for  operating  when  each  man's  stock  is 
employed  for  the  same  period  of  time'?  What  is  the  rule  for  cancel- 
ing 1  When  all  the  shares  are  not  continued  for  the  same  period  of 
time,  on  what  will  each  man's  share  of  the  profit  and  loss  depend'? 
What  is  the  rule  1 


COMMERCIAL    EXCHANGE. 

Under  this  rule  are  included  the  operations  of  purchasing 
goods  in  one  country,  in  the  currency  of  that  country,  and  sell- 
ing them  at  wholesale  or  retail  in  the  currency  of  another  coun- 
try, so  as  to  give  or  lose  some  required  per  cent. 


TABLE  OF  GOLD  COINS. 


NAMES  OF  COINS. 


VALUE. 


Austrian  Dominions. 
Souverein,  -        -      $3.37  7 

Double  Ducat,      -        -        4.58  9 
Hungarian  Ducat,       -        2.29  6 

Bavaria . 

Carolin,       -        -        -  4.95  7 

Max  d'or  or  Maximilian,  3.31  8 

Ducat,         -       -       -  2.27  5 


NAMES  OF  COINS.  VALUE. 

Berne. 

Ducat,  double  in  pro- 
portion.   -        -  $1.98  6 
Pistole,         -        -        -        5.54  2 

Brazil. 

Johannes,  half  in  propor- 
tion,-       -        -        -      17.06  4 
Dobraon,     -       -       -      32.70  6 


224 


COMMERCIAL  EXCHANGE. 


NAMES  OF  COINS. 


VALUE. 


Dobra,          -        -        -    $17.30  1 
Moidore,  half  in  propor- 
tion,         -        -        -        6.55  7 
Crusade,      ...          63  5 

Brunswick. 

Pistole,  double  in  propor- 
tion,        -        -         -        4.54  8 
Ducat,         -        -         -        2.23 

Cologne, 
Ducat,         -        -       -       2.26  7 

Colombia. 
Doubloon,   -        -        -      15.53  5 

Denmark. 
Ducat,  Current,  -        -        1.81  2 


Ducat,  Specie, 
Christian  d'or,      - 


2.26  7 
4.02  1 


East  India. 

Rupee,  Bombay,          -  7.09  6 

Rupee,  Madras,  -        -  7.11 

Pagoda,  Star,       -        -  1.79  8 

England. 

Guinea,  half  in  propor- 
tion,         -        -        -  5.07  5 

Sovereign,  half  in  propor- 
tion,         -        -        -  4.84  6 

Seven  shilling  piece,  -  1.69  8 

France. 

Louis,  coined  before  1786,  4.84  jj 

Double  Louis,  before  1786,  9.69  J 

Louis,  coined  since  1786,  4.57  « 

Double  Louis,  since  1786,  9.15  3 

Napoleon,  or  20  francs,  3.85  1 
Double  Napoleon,  or  40 

francs,      -        -        -  7.70 

Frankfort  on  the  Main. 

Ducat,          -       -        -  2.27  9 

Geneva. 

Pistole,  old,          -        -  3.98  5 


Pistole,  new, 

Hamburg. 

Ducat,  double  in  propor- 
tion,        - 


3.44  4 


2.27  9 


NAMES  OF  COINS.  VALUE. 

Genoa. 
Sequin,        -  $2.30  2 

Hanover. 

Double  George  d'or,   -       7.87  9 
Ducat,          -        -        -        2.29  6 
Gold  Florin,  double  in  pro- 
portion,   -        -        -        1.67 


Holland. 

Double  Ryder,     - 

Ryder, 

Ducat, 

Ten  Guilder  Piece, 


Malta. 


Double  Louis, 

Louis, 

Demi-Louis, 


Doubloons, 


Mexico. 


Milan. 


12.20  5 
6.04  3 
2.27  5 
4.03  4 


9.27  8 
4.65  2 
2.33  6 


15.53  5 


Sequin,        -        -        -        2.29 
Dopia  or  Pistole,          -        3.80  7 
Forty  Livre  Piece,       -        7.74  2 

Naples. 

Six  Ducat  Piece,  1783,  5.24  9 
Two  do.  or  Sequin,  1762,  1.59  1 
Three  do.  or  Oncetta,  1818,  2.49 

Netherlands. 

Gold  Lion,  or  14  Florin 

Piece,       -        -        -        5.04  6 
Ten  Florin  Piece,       -        4.01  9 

Parma. 

Quadruple  Pistole,       -  16.62  8 

Pistole,  or  Dappia,  1787,  4.19  4 

do.            do.      1796,  4.13  5 

Maria  Theresa,      1818,  3.86  1 

Piedmont. 

Pistole,  coined  since  1785,    5.41  1 
Sequin,  half  in  proportion,  2.28 
Caflino,  coined  since  1785, 27.34 
Piece  of  20  Francs,  or 
Marengo,  3.56  4 


COMMERCIAL  EXCHANGE. 


225 


NAMES  OF  COINS. 


VALUE.       NAMES  OP   COINS. 


Poland. 


Ducat, 


$2.27 


Portugal. 

Dobraon,  -  -  -  32.70  8 
Dobra,  -  -  -  17.30  1 
Johannes,  -  -  -  17.06  4 
Moidore,  half  in  propor- 
tion, -  -  6.55  7 
Piece  of  16  Testoons,  1600 

Rees,  -  -  -  2.12  1 
OldCrusado,  or  400  Rees,  58  8 
New  Crusado,  or  480  Rees,  63  5 

Milree,  coined  in  1755,  73 

Prussia. 

Ducal,  1748,        -        -  2.27  9 

Ducat,  1787,        -        -  2.26  7 

Frederick,  double,  1769,  7.97  5 

Frederick,  double,  1800,  7.95  1 

Frederick,  single,  1778,  3.99  7 

Frederick,  single,  1800,  3.97  5 

Rome. 

Sequin,  coined  since  1760,  2.25  1 

Scudo  of  Republic,      -  15.81  1 

Russia. 

Ducat  of  1796,    -       -  2.29  7 

Ducat  of  1763,     -        -  2.26  7 

Gold  Ruble  of  1756,    -  96  7 

Gold  Ruble  of  1799,    -  73  7 

Gold  Pollen  of  1777,   -  35  5 

Imperial  of  1801,         -  7.82  9 

Half  Imperial  of  1801,  3.91  8 

Half  Imperial  of  1818,  3.93  3 

Sardinia. 

Carlino,    half  in  propor- 
tion,         -        -        -  9.47  2 

Saxnny. 

Ducat  of  1784,    -     .  -  2.26  7 

Ducat  of  1797,    -        -  2.27  9 

Augustus  of  1754,        -  3.92  5 

Augustus  of  1764,       -  3.97  4 


Sicily. 

Ounce  of  1751,    -        -      $2.50  4 
Double  Ounce,    -        -        5.04  4 

Spain. 

Doubloons  of  1772,  parts 

in  proportion.  -        -      16.02  8 

Doubloon,    -        -        -      15.53  5 

Pistole,        -        -        -        3.88  4 

Coronilla,  Gold  Dollar,  or 

Vintern,  1801,  98  3 

Sweden. 
Ducat,         -       -       -       2.23  5 

Switzerland. 

Pistole  of  the  Helvetic  Re- 
public,  -        -        -          4.56 

Treves.  • 
Ducat,         -       -       -       2.26  7 

Turkey. 

Sequin  fonducli  of  Con- 
stantinople,      -        -        1.86  8 
Sequin  fonducli  of  Con- 
stantinople, 1789,      -        1.84  8 
Half  Misseir,  1818,     -  52  1 

Sequin  fonducli,  -        1.83 

Yeermeeblekblek,        -        3.02  8 

Tuscany. 

Zechino  or  Sequin,     -        2.31  8 
Ruspone  of  the  Kingdom 
of  Etruria,       -       -        6.93  8 

Venice. 

Zechino  or  Sequin,  shares 
in  proportion,  -        -        2.31 

Wirteviberg. 

Carolin,       -        -        -        4.89  8 
Ducat,  -        -        2.23  5 

Zurich. 

Ducat,  double  and  half  in 
proportion,       -        -       2.26  7 


226 


COMMERCIAL  EXCHANGE. 


TABLE  OF  UNCOINED  AND  SILVER  MONEY. 


English,  Currency. 
1£.  sterling,  before  1832 

=$4.444.  Since  1832=  $4.80 
Or  prior  to  1832,  9£.= 

$40.  Si  nee  1832, 5  £.=  24.00 

1  English  crown,          -  1.10 

OrlOcrowns=            -  11.00 

1  English  shilling,        -  22  2 

1  pistareen,           -  20 

Or  5  pistareens=         -  1.00 

1  English  penny,  0185 

Irish  Currency. 

1  £.  Irish,    -        -        -  4.10 

Or  10  £.=           -        -  41.00 

1  shilling,    ...  20  5 

1  penny,  01  7 

French  Currency. 

1  French  crown,          -  1.10 

Or  10  crowns=  -        -  11.00 

1  five  franc  piece,        -  93 

1  franc,        ...  J8  6 

1  Decimes,  0186 

Spanish  Currency. 

1  Spanish  dollar,          -  1.00 

1  real,  new  plate,         -  10 

1  real  vellon,       -  05 


Currencies  of  other  Nations. 
millree  of  Portugal,        $1.24 

Russian  silver  ruble,  75 

rix  dollar  of  Sweden,  1.00 

Russian  rix  dollar  66  6 

Or  3  rix  dollars=        -  2.00 
Danish  rigsbank  dollar,       50 

silver  ducat  of  Naples,  80 

Or  5  ducats=       -        -  4.00 

scudo  of  Sicily,        -  96 

oncia  of  Sicily,         -  2.40 

pezza  of  Leghorn,    -  90 

pezza  of  Genoa,        -  89 

flourin  of  Trieste,    -  48 

rix  dollar  of  Trieste,  96 

Roman  crown,           -  1.00 

gold  crown  of  Rome,  1.53 

Maltese  scudo,  40 

rupee  of  Bengal,       -  55  5 

rupee  of  Bombay,     -  50 

pagoda  of  Madras    -  1.80 

tale  of  Canton,          -  1.48 

Japanese  tale,  75 

dollar  of  Sumatra,    -  1.10 

tale  of  Sumatra,       -  4.16 

florin  of  Java,  40 
mark  banco  of  Hamburg,   33  3 

guilder  of  Amsterdam,  40 


To  reduce  a  foreign  currency  to  federal  money,  multiply  the 
given  money  of  that  currency  by  the  value  of  its  unit  in  federal 
money,  as  given  in  the  preceding  tables. 

If,  however,  it  be  required  to  find  how  goods,  the  value  of 
which  is  given  in  a  foreign  currency,  must  be  sold  in  federal 
money,  to  gain  or  lose  a  given  per  cent.,  first  change  the  for- 
eign currency  to  federal  money,  and  calculate  the  gain  or  loss  on 
this  by  the  rules  already  giv^n.  The  retail  price  of  any  de- 
nomination, as  cwt.,  lb.,  yd.,  gal.,  &c.,  may  then  be  found  by 
division. 

Ex.  1.  Purchased  in  London  360  yards  of  broadcloth,  which 
cost  me,  including  transportation,  300  £>.  sterling  ;  how  must  I 
sell  the  same  per  yard,  in  federal  money,  to  make  a  profit  of 
20  per  cent.  ? 


COMMERCIAL  EXCHANGE.  227 

Solution:  300  £.  x  40=  12000;  and  12000-h9=$1333|3 
the  value  of  300  £.  in  federal  money. 

Then,  $1333|x  1.20=$1600,  the  value  in  federal  money 
increased  by  the  gain  per  cent.  ;  therefore,  $1600-^-360  = 
$4.444,  -f  the  required  price  of  one  yard. 

If  the  wholesale  price  only  be  required,  it  is  only  necessary 
to  omit  the  last  step.  Thus  the  $1600  is  the  wholesale  price 
of  the  cloth  given  in  the  preceding  sum. 

To  solve  sums  like  the  preceding  by  canceling  : 

RULE.  —  Place  the  whole  cost  in  the  given  currency  first  above, 
and  (if  the  retail  price  be  required)  the  number  expressing  the 
quantity  procured  for  that  price,  first  below,  a  horizontal  line. 
Write  next  above  the  line  the  value  of  a  unit  of  the  given  cur- 
rency, in  federal  money.  And,  lastly,  to  increase  or  diminish 
the  price  by  the  required  per  cent.,  place  100  below  the  line,  and 
100  increased  by  the  per  cent,  to  be  gained,  or  diminished  by  the 
per  cent,  to  be  lost,  above  the  same. 

Note  1st.  —  If  the  wholesale  price  be  required,  the  number 
expressing  the  whole  quantity  (by  the  preceding  rule,  placed 
below  the  line)  must  be  rejected. 

Note  2d.  —  Whenever  in  the  preceding  tables  the  value  of  a 
number  of  units  of  foreign  currency  is  given  in  federal  money, 
place  that  number  below  the  line,  and  its  federal  value  above  the 
sum.  (See  -£'s  sterling.) 

The  preceding  sum  stated  for  canceling  : 

300.  40.  120 

360.    9.  100' 

To  understand  the  reason  of  the  middle  terms,  see  Note  2d, 
and  <£'s  sterling  in  the  preceding  tables. 

40 
SBtt.  4ft.  121ft 


9.    iM' 
9     8 
And  40  4-  9  =  $4.  444,  -f  the  answer,  the  same  as  before. 

2.  Purchased  in  London  350  yards  of  sheeting  for  75  £.,  and 
paid  12  £,.  for  its  transportion  to  New  York  city  ;  how  must  I 
retail  the  same  in  federal  money,  to  gain  15  per  cent,  on  the 
first  cost?  Ans.  $1.27.+ 


228  COMMERCIAL  EXCHANGE. 

Statement,  350    9'  100-    The  87  in  the  statement:=75.£. 

3.  Received  from  London  470  yards  of  dimity,  which,  in- 
cluding transportation,  cost  me  65  £. ;  sold  the  same  by  the 
yard  so  as  to  gain  30  per  cent,  on  the  first  cost ;  how  did  I 
sell  it  ?     Ans.  $0.799 -f  per  yard. 

4.  Received  from  Dublin  600  yards  of  Irish  linen,  the  whole 
cost  of  which  was  75  £>.  Irish  currency  ;  how  must  I  retail  the 
same  in  federal  money,  to  gain  12|  per  cent.  ?    Ans.  $0.576+ 
per  yard. 

5.  My  agent  in  Dublin  has  forwarded  to  me  900  yards  of 
linen  ;  whole  cost  60  £.  Irish  currency  ;  how  must  I  retail  the 
same  in  federal  money,  to  gain  15  per  cent.  1    Ans.  $0.314.+ 

6.  I  have  in  my  store  120  yards  of  broadcloth,  forwarded 
me  by  my  agent  in  Paris,  which  cost  me,  including  transpor- 
tation, 325  crowns  ;  how  must  I  sell  the  same  in  federal  money, 
to  gain  16  per  cent.  ?     Ans.  $3.455+  per  yard. 

7.  Received  680  yards  of  silk  from  Paris,  for  which  I  paid 
560  crowns  ;  expenses  of  transportation,  12  crowns;  how  must 
I  sell  the  same  in  federal  money,  to  gain  30  per  cent.  ?     Ans. 
$1.20+  per  yard. 

8.  Received  from  Madrid  6  hogsheads  of  wine,  each  con- 
taining 63  gallons,  for  which  my  agent  paid  188  Spanish  dol- 
lars ;  how  must  I  sell  the  same  per  gallon,  to  gain  12^  per  cent.  ? 
Ans.  $0.559.+ 

9.  I  have  on  hand  a  bale  of  silk,  containing  174  yards,  which 
I  received  from  Cadiz,  at  a  cost,  including  transportation,  of 
140  piastres  or  Spanish  dollars  ;  how  must  I  sell  the  same  per 
yard,  to  gain  5  per  cent.  ?     Ans.  $0.844.+ 

10.  Received  from  Oporto  3  hogsheads  of  port  wine,  con- 
taining 63  gallons   each ;    cost,  including  transportation,   30 
milrees  per   hogshead ;  how  must  I  retail   the  same  by  the 
gallon,  to  gain  25  per  cent.?     Ans,  $0.738.+ 

1 1 .  How  must  I  sell  broadcloth  by  the  yard,  in  federal  money, 
of  which  3  pieces,  each  containing  35  yards,  cost  me  135  £. 
sterling,  to  gain  30  per  cent.  1     Ans.  $7.428.+ 

12.  Received  from  A.  B.,  Dublin,  560  yards  of  linen ;  whole 
cost  90  £.  Irish  currency  ;  what  must  be  the  retail  price,   in 
federal  money,  to  gain  15  per  cent.  ?     Ans.  $0.757+   por 
yard. 

13.  Received  from  the  same  600  yards  of  muslin,worth  56  £. ; 
how  must  I  sell  the  whole  quantity  in  federal  money,  to  gain 
5  per  cent.  ?     Ans.  $241.08. 


COMMERCIAL  EXCHANGE.  229 

14.  Consigned  to  my  agent,  J.  Jones,  of  London,  300  barrels 
of  flour,  for  which  I  paid  $1500  ;  how  many  pounds  sterling 
ought  he  lo  receive  for  the  same,  to  gain  10  per  cent.,  the  ex- 
pense of  transportation  being  $50?     Ans.  383  £.  12s.  6d. 

15.  Received  of  my  agent  in  London,  J.  Jones,  2510  gallons 
of  Madeira  wine,  which  cost  me,  per  invoice,  1640  j£.  sterling  ; 
but  it  being  of  an  inferior  quality,  I   am  willing  to  lose  5  per 
cent,  on  the  cost ;  what  must  be  the  price  per  gallon,  in  federal 
money?     Ans.  $2.758.+ 

16.  Three  men  trading  in  company,  received  from  France 
1200  bottles  of  champagne,  for  which  they  paid  600  French 
guineas,  each  $4.60  ;  how  must  they  sell  the  same  per  bottle, 
in  federal  money,  to  gain  40  per  cent.,  and  what  will  be  each 
man's  gain  per  bottle  ?     Ans.  $3.22  per  bottle  ;  each  man's 
gain  per  bottle,  $0.306.+ 

17.  Received  300  ells  of  cloth  from  Hamburgh,  which  cost 
me  1 500  mark  bancos  ;  how  must  the  same  be  sold  in  federal 
money,  by  the  yard,  to  gain  12-2-  Per  cent.,  the  ell  Hamburgh 
being  2£  qr.  ?     Ans.  $1.17.+ 

18.  New  York,  Jan.  6,  1838.     This  day  received  from  Am- 
sterdam, 600  yards  of  carpeting  ;  whole  cost,  2400  guilders. 
Required  the  retail  price  in  federal  money,  to  gain  20  per  cent. 
Ans.  $  1  92  per  yard. 

19.  Shipped  to  London  380  barrels  of  flour,  which  cost  me, 
including  transportation,  $6  per  barrel.     How  many  English 
crowns  must  I  receive   for  the  whole  quantity,  to  gain  10  per 
cent.     Ans.  2280  crowns. 

20.  Shipped  to   Dublin   3000  bushels   of  flax-seed,  which 
cost  me  $2500.  How  many  pounds,  Irish  currency,  must  I  re- 
ceive for  the  whole  quantity,  to  gain  5  per  cent.  ?  Ans.  640  £. 
4s.  10  d.  2qr.+ 

21.  Boston,  Jan.   16,   1835.     This  day  received  from  my 
agent  at  Lisbon,    16  hogsheads   of  wine,   each  65  gallons  ; 
whole   cost,  640  milrees.     The   whole  being  of  an   inferior 
quality,  I  am  willing  to  lose  6  per   cent,   on  the  cost.     How 
much  in  federal  money  must  I   charge  per  gallon?      Ans. 
$0  72,  nearly. 

22.  New  York,  Sept.  6, 1835.     This  day  received  from  A. 
B.,  London,  1200  yards  of  superfine  broadcloth  ;  whole  cost, 
1500.6.     How  must  1   sell  the   same  in  federal  money,  at 
wholesale,  to  realize  a  profit  of  20  per  cent.  ?     Ans.  $8000. 

23.  Received  from  Russia  a  quantity  of  fur  ;  whole  cost,  900 
silver  rubles.     What  must  be  the  wholesale  price  in  federal 

20 


\ 

230  TARE  AND  TRET. 

money,  to  make  an  advance  of  15  per  cent,  on  the  cost  ?  Ans. 
$776.25. 

24.  Boston,  Feb.  26,  1836.     This  day  received  from  my 
agent  in  Paris,  24  hogsheads  French  wine,  each  60  gallons ; 
whole  cost,  288  guineas.     How  must  the  same  be  retailed  by 
the   quart,  in  federal  money,  to  gain  12^  per  cent.?      Ans. 
$0.25.8.+ 

25.  Shipped  to  my  agent  in  London,  650  barrels  of  flour  ; 
whole  cost,   $3600.     How  many  pounds  sterling  must  I  re- 
ceive for  the  same,  to  gain  10  per  cent,  on  the  cost  ?     Ans. 

891  £. 

/A  tfv^a4L4A''— ^ 

QUESTIONS. — What  operations  are  included  under  this  rule  1  How 
is  a  foreign  currency  reduced  to  federal  money  1  How  is  the  opera- 
tion performed,  when  it  is  required  to  find  how  goods,  the  value  of 
which  is  given  in  a  foreign  currency,  must  be  sold  to  gain  or  lose  a 
certain  per  cent,  in  federal  money  1  What  is  the  rule  for  canceling  1 
What  is  Note  1st  1  What  is  Note  2d  1 


TARE    AND    TRET. 

We  come  now  to  consider  the  allowances  to  be  made  in  the 
purchase  of  goods  by  weight.  The  following  particulars 
require  to  be  first  noticed : 

Gross  Weight  is  the  whole  weight  of  the  goods  purchased, 
including  that  of  the  box,  barrel,  bag,  &c.  containing  them. 

Draft  is  a  deduction  from  the  gross  weight  made  in  favor  of 
the  buyer. 

Tare  is  an  allowance  made  for  cask,  box,  barrel,  &c.  con- 
taining the  goods  ;  and  may  be  either  a  certain  deduction  from 
the  whole  quantity,  or  so  much  per  box,  &c. 

Tret  is  an  allowance  of  4  Ib.  for  every  104lb.  made  for  the 
dust,  &c. 

Suttle  is  what  remains  after  some  of  the  preceding  allow- 
ances have  been  made. 

Net  Weight  is  what  remains  after  all  the  deductions  have 
been  made. 


TARE  AND  TRET.  ,  231 

Ex.  1.  Bought  a  hogshead  of  sugar,  weighing  7  cwt.  2qr. 
26  lb.,  tare  on  the  whole,  3  qr.  24  Ib.     What  is  the  net  weight  ? 

cwt.  qr.     lb. 

7     2     26 

3     24 


632  Net  weight. 

2.  What  is  the  net  weight   of  12  casks  of  raisins,  each 
weighing  2  cwt.  2  qr.  14  lb.,  tare  per  cask,  12  lb.  ? 

2  cwt.  2  qr.  141b.  X  12  =  31  cwt.  2  qr.,  the  gross  weight.  12 
X  12  =  144  lb.  =  1  cwt.  1  qr.  4  lb.  ;  and  31  cwt.  2  qr.  —  1  cwt. 
1  qr.  4  lb.  =  30cwt.  Oqr.  24  lb.  Ans. 

3.  What   is  the  net   weight  of  6  casks   of  prunes,  each 
weighing  3  cwt.  2  qr.  lOlb.,  tare  20 lb.  per  cask?     Ans.  20 
cwt.  1  qr.  24  lb. 

4.  What  is  the  net  weight  of  44  cwt.  gross,  if  14  lb.  per  cwt. 
be  allowed  for  tare  1     44 x  14=616  lb.  =  5  cwt.  2qr.,  and  44 
cwt.  —  5  cwt.  2  qr.  =  38  cwt.  2  qr.  Ans. 

Or  the  solution  may  be  effected  by  canceling,  by  the  follow- 
ing rule  : 

RULE. — Place  the  whole  gross  weight  first  above  a  horizontal 
line.  Then  place  112  lb.  below  the  line,  with  112  diminished  by 
the  tare  per  cwt.,  standing  directly  above  it.  Cancel,  tyc. 

The  above  sum  solved  by  this  rule  : 

11    U  7 
112-14=98.     itj.     Canceled,  *t«-f 

S&  2 
and  11  x7=77,  and  77-^2  =  38  cwt.  2  qr. 

5.  Bought  84  cwt.  of  sugar.     What  is  the  net  weight,  if  20 
lb.  per  cwt.  be  allowed  for  tare  ?     Ans.  69  cwt. 

6.  Bought  9  hogsheads  of  sugar,  each  weighing  8  cwt.  2 
qr.      From  this,  a  deduction  of  161b.  per  cwt.  was  made  for 
tare.     What  was  the  net  weight?     Ans.  65  cwt.  2  qr.  8  lb. 

Note  1st. — When  the  price  per  cwt.  or  per  lb.  is  given,  the 
reduction  for  tare  and  tret  may  be  made,  and  the  whole  cost  as- 
certained by  a  single  statement,  as  may  be  seen  from  the  follow- 
ing example : 


232  «  TARE  AND  TRET. 

7.  What  is  the  value  of  8  hogsheads  of  sugar,  each  weigh- 
ing 12  cwt.  gross,  tare  12  Ib.  per  cwt.,  at  $8.50  per  cwt.  ? 


Statement,  .      Canceled, 

7 

and8.50xlOOx3x2^7=:$728.57,  Ans. 

8.  Bought  32  casks   of  figs,  each  weighing  2  cwt.  2qr.,  at 
a  deduction  of  181b.  per  cwt.  for  tare.     What  did  the  whole 
cost  me,  at  $4  per  cwt.  net  weight  ?     Ans.  $268.57. 

9.  Bought  15  cwt.  of  sugar  at  $6.50  per  cwt.  net  weight. 
Reduction  for  tare,  12  Ib.  per  cwt.  ;  tret,  4  Ib.  per  104  Ib.  How 
must  I  sell  the  whole,  to  gain  20  per  cent,  on  the  first  cost,  and 
how  must  I  retail  it,  to  gain  the  same  per  cent.  ?    Ans.  Whole- 
sale price,  $100.446  ;  retail  price,  $0.059,  nearly. 

To  effect  all  the  reductions  and  the  gain  per  cent,  of  the 
preceding  sum,  a  single  statement  only  is  required  ;  thus  : 

15.  6.50.  100.  100.  120 
11  2.  104.  100. 

10.  Bought  32  chests  of  tea,  each  weighing  4  cwt.  2qr.  at 
$49  per  cwt.  net  weight;  tare,  12  Ib.  per  cwt.  ;  tret,  41b.  per 
104lb.     How  must  I  sell  the  whole  quantity  to  gain  20  per 
cent.,  and  how  must  the  same  be  retailed,  to  gain  the  same  ? 
Ans.  Wholesale  price,  $7269.23  ;  retail  price,  $0.525. 

11.  Bought  742  Ib.  of  wool,  and  was  allowed  a  deduction  of 
5  per  cent,  from  the  gross  weight  for  dust,  &c.     For  the  net 
weight  I  paid  9  s.  New  York  currency,  per  pound,  and  was  al- 
lowed a  deduction  of  6  per  cent,  on  the  whole  cost  for  ready 
money.     1  then  sold  the   same   so  as  to  realize  a  profit  of  20 
per  cent,  on  the  money  I  advanced.     How  much  did  I  receive 
for  the  whole  ?     Ans.  $900. 

12.  Purchased  5  cwt.  of  sugar;  tare  allowed,  8  Ib.  per  cwt. 
For  the  net  weight  I   paid   6  d.   New  York  currency,  per  Ib. 
How  must  I  sell  the  whole  quantity  to  gain  20  per  cent.,  and 
how  must  the  same  be  retailed,  to  gain  the  same  per  cent.  ? 
Ans.  Wholesale  price,  $39  ;  retail  price,  $0.07£  per  pound. 

13.  Purchased  12  bags  of  coffee,  each  weighing  96  Ib.  ;  tare 
per  bag,  6  Ib.     What  was  the  whole  cost  at  30  cents  per  Ib. 
and  the  retail  price,  to  gain  25  per  cent.  ?     Ans.  $324  whole 
cost  ;  $0.37£  retail  price. 

14.  How  much  will  8  hogsheads  of  sugar,  each  weighing 


TARE  AND  TRET.  233 

8  cwt.  3  qr.,  cost  at  $9  per  cwt.  if  a  deduction  of  12  Ib.  per 
cwt.  be  allowed  for  tare ;  and  what  will  be  received  for  the 
whole,  if  it  be  sold  at  an  advance  of  30  per  cent.  ?  Ans. 
$562.50  cost,  and  $731.25  received. 

15.  What   is  the   net   weight   of  3    tierces  of  rice,  each 
weighing  4  cwt.  3  qr.  gross  ;  the  tare  allowed,  16  Ib.  per  cwt. ; 
tret,  4lb.  per  104  Ib.  ?  Ans.  11  cwt.  2qr.  27lb.+ 

16.  What  is  the  cost  of  15  chests  of  tea,   each  containing 
1401b.  gross,  at  5  s.  New  York  currency,  per  Ib.  ;  tare,  16  Ib. 
per  cwt.     Ans.  $1125. 

17.  Bought  12  hogsheads  of  sugar,  each  weighing  10  cwt. 
2qr.  at  $9  per   cwt.   net  weight.      Deduction  for  tare,  12  Ib. 
per  cwt.    How  much  must  I  receive  for  the  whole  quantity,  to 
gain  10 per  cent,  on  the  cost?    Ans.  $1113.75. 

18.  Bought  16  firkins  of  butter,  each  weighing  108  Ib.  ;  re- 
duction for  tare,  81b.  per  cwt.     Paid  15  pence,  New  England 
currency,  per  pound.     What  did  it  cost  me  ;   and  what  must  be 
the  wholesale,  and  what  the  retail  price,  to  gain  20  per  cent,  on 
the  first  cost  1  Ans.  Whole  cost,  $334.285  ;-j-  wholesale  price, 
$401.142;  retail  price,  $0.25. 

19.  Bought  18  cwt.  of  sugar,  at   $12  per  cwt.  net  weight ; 
tare,  16  Ib.  per  cwt.     How  must  I  sell  the  same  per  Ib.  to  gain 
12  per  cent,  on  the  first  cost  1     Ans.  $0.12. 

20.  Purchased  in  London,  16  cwt.  of  tea  at  28  £.  sterling 
per  cwt.  net  weight ;  tare,  12  Ib.  per  cwt.     How  much  must 
I  receive  in  federal  money,  for  the  whole  quantity,  to  realize  a 
profit  of  12   per  cent.,  and  what  retail  price  will  allow  the 
same  profit  ?     Ans.  Wholesale  price,  $1991.11  ;  retail  price, 
$1.24. 

Q.UESTIONS. — What  is  Tare  and   Tret  1    What  is  gross  weight  7 
What  is    draft1?    What  is  tare'?    What  is  tret"?    What  is  suttle  ? 
What  is  net  weight  1    What  is  the  rule  7    What  is  the  note  7 
20* 


234  EQUATION  OF  PAYMENTS. 


EQUATION    OF    PAYMENTS. 

Equation  of  Payments  is  the  method  of  finding  a  mean 
time  for  the  payment  of  several  debts  due  at  different  periods 
of  time. 

Ex.  1.  I  owe  a  friend  $380,  to  be  paid  as  follows,  viz, 
$100  in  6  months;  $120  in  7  months;  and  $160  in  10 
months.  If  I  pay  the  whole  at  once,  at  what  time  must  the 
payment  be  made,  so  that  neither  I  nor  my  friend  shall  lose  in- 
terest ? 

SOLUTION. 

The  interest  of  $  100  for  6  mo.  —the  interest  of  $  1  for  600  mo. 
The  interest  of  $120  for  7  mo.^the  interest  of  $1  for  840  mo. 
The  interest  of  $160  for  10  mo. = the  interest  of  $1  for  1600  mo. 


Amt.  of  pay'ts,  $380  3040  mo, 

3040 — the  months  requisite  for  $1  to  gain  as  much  interest 
as  $380  would  gain  in  the  required  time.  Therefore,  $380  : 
$1  :  :  3040  months :  to  the  required  time,  viz.  8  months. 

The  600,  840,  and  1600  months  are  obviously  obtained  by 
multiplying  the  several  payments  by  the  time  which  must 
elapse  before  they  severally  become  due. 

We  therefore  have  the  following  rule : 

RULE. — Multiply  each  payment  by  the  time  which  must  elapse 
before  it  becomes  due,  and  divide  the  sum  of  the  products  by 
the  sum  of  the  payments. 

2.  A.   owes  me  $50,  payable  in  4  months  ;  $100,  payable 
in  10  months  ;  and  $150,  payable  in  16  months.     In  what 
time  must  he  pay  the  whole,  so  that  neither  shall  lose  interest? 
Ans.  12  months. 

3.  What  is  the  equated  time  of  payment  for  the  three  fol- 
lowing sums,  viz.  $500,  payable  in  3  years  ;  $400,  payable  in 
4  years  ;  and  $600,  payable  in  5  years  ?     Ans.  4^  years. 

4.  What  is  the  equated  time  of  payment  for  the  three  fol- 
lowing sums,  viz.  $50,  payable  in  4  months  ;  $75,  payable  in 
6  months  ;  and  $100,  payable  in  7  months  ?     Ans.  6  months. 


DUODECIMALS.  235 

5.  A.  owes  B.  $400  ;  of  which  $80  is  payable  in  6  months; 
$120  in  ten  months ;  and  the  remainder  in  1  year  and  8  months. 
What  is  the  equated  time  of  payment  ?   Ans.  1  year,  2  months, 
and  6  days. 

6.  A  merchant  has  a  certain  sum  of  money  due  him,  of 
which  £  is  payable  in  2  months  ;  3  in  4  months  ;  and  the  re- 
mainder in  6  months.     What  is  the  equated  time  for  the  pay- 
ment of  the  whole  ?     Ans.  4-J  months. 

7.  I  owe  four  sums  of  money,  payable  as  follows,  viz.  $60 
payable  in  9  months  ;  $80  in  10  months  ;  $50  in  11  months  ; 
and  $60  in  12  months.     At  what  time  may  I  pay  the  whole, 
without  loss  ?    Ans.  lO^J  months. 

8.  A  person  owes  a  debt  of  $2000,- payable  in  7  months,  of 
which  he  proposes  to  pay   $600  down,  on  condition  that  the 
remainder  be  allowed  to  remain  unpaid  an  adequate  term  of 
time.     In  what  time  ought  it  to  be  paid  ?     Ans.  10  months. 

QUESTIONS. — What  is  Equation  of  Payments  1    What  is  the  rule  ? 


DUODECIMALS. 

Duodecimals  are  fractions  of  afoot.  The  unit  or  foot  is 
first  supposed  to  be  divided  into  12  equal  parts,  called  inches 
or  primes,  and  marked'.  Each  inch  or  prime  is  then  divided 
into  12  equal  parts,  called  seconds,  and  marked  ".  Each  second 
is  again  divided  in  like  manner,  and  each  part  obtained  by  this 
division  is  called  a  third,  and  marked  '".  A  foot  is  therefore 
divided  duodecimally,  when  it  is  separated  into  12  equal  parts, 
or  when  the  several  divisions  sustain  a  twelve-fold  relation  to 
each  other.  This  relation  may  be  thus  illustrated: 

1' inch  or  prime  is y^  of  a  foot. 

1  "second  is  y1^  of  Jj, ^  of  a  foot. 

1"'  third  is  TJ2  of  -^  of  y1^,  -     -     -     -  yJ^  of  a  foot. 

1 ""  fourth  is  y1^  of  yL  of  -jV,  of  -fa,  -  -  yo^-g  of  a  foot. 
I'""  fifth  is  TT2  of  y1^  of  yL  of  yL-  of  -jL,  -  2t8\32  of  a  foot. 
I"""  sixthisy^of  y^ofyLofy^ofy^ofyL,  ^.1^- Of  a  foot,&c. 


236  MULTIPLICATION   OF    DUODECIMALS. 

The  marks  ',  ",  "',  "",  &c.  are  the  indices  of  the  several  de- 
nominations. 

TABLE   OF  DENOMINATIONS. 

12"""  sixths  make I'1'"  fifth. 

12"'"  fifths  make I""  fourth. 

12""  fourths  make I'"  third. 

12'"  thirds  make 1"  second. 

12"  seconds  make 1'  prime  or  inch. 

12'  inches  or  primes  make  -     -     -  1  foot. 

Duodecimals  may  be  added  or  subtracted  in  the  same  man- 
ner as  compound  numbers ;  the  denominations  decreasing  or  in- 
creasing in  the  constant  ratio  of  12.  These  operations  are  so 
simple  that  it  is  unnecessary  to  introduce  any  examples.  The 
principle  is  the  same  as  in  Compound  Addition  and  Subtraction. 


MULTIPLICATION  OF  DUODECIMALS. 

The  greatest  difficulty  the  scholar  will  here  encounter, 
will  be  to  determine  the  denomination  of  the  product  of 
any  two  denominations.  Suppose  it  be  required  to  multiply 
6'  inches  or  primes  by  4'  inches,  their  product  is  obviously  24, 
but  of  what  denomination  is  it  ?  By  recurring  to  the  preceding 
tables,  it  will  be  seen  that  6'  inches =T%  of  a  foot,  and  4'  inches 
=:T4^  of  a  foot ;  and  T^-  X  yj—  -f^,  that" is,  24".  Hence,  inches 
multiplied  by  inches  produce  seconds.  Again,  let  it  be  re- 
quired to  multiply  9''  seconds  by  3'  inches.  9  seconds  =  i%-±, 
and  3  inches =-f% ;  therefore,  yl^  x  Ty=T?y8  > or  27'".  Lastly, 
what  is  the  product  of  I'1  seconds  multiplied  by  9"  seconds. 
7"  seconds — xJ^,  and  9"  seconds  =  3-^,  and  T77XTT4"— 
2o6T33ir>  or  63////  fourths.  From  the  above  we  draw  the  follow- 
ing conclusions,  viz.  that  feet  multiplied  by  feet  produce  square 
feet ;  feet  multiplied  by  inches  produce  inches ;  inches  or  primes 
multiplied  by  inches,  give  seconds;  seconds  multiplied  by  seconds, 


MULTIPLICATION  OF  DUODECIMALS.  237 

give  fourths  ;  seconds  multiplied  by  thirds,  give  fifths,  fyc.  ; 
that  is,  the  product  of  any  two  denominations  will  always  be  of 
the  denomination  expressed  by  the  sum  of  their  indices. 

RULE. — Begin  with  the  lowest  denomination  of  the  multipli- 
cand, and  multiply  it  by  the  highest  denomination  of  the  multi- 
plier, and  place  each  term  of  the  product  according  to  its  respec- 
tive value.  Multiply  in  the  same  manner  by  each  remaining  de- 
nomination of  the  multiplier,  and  place  the  product  of  each  suc- 
ceeding multiplication  one  or  more  places  farther  to  the  right, 
according  to  the  denomination.  The  several  products  thus  ob- 
tained, when  added  together,  will  give  the  required  product. 

Note. — It  will  be  remembered  to  carry  by  12  in  all  cases. 
Ex.  1.  Multiply  4  feet  4  inches,  by  4  feet  4  inches. 

OPERATION. 

ft.     in. 

4     4'  Beginning  with  the  4  feet  in  the  mul- 

4     4'  tiplier,  we  say  4  times  4'  are  16'r=l  ft. 

4'.     The  4'  is  set  down  and  the  1  ft. 

17  4'  carried  to  the  next  product,  making  it 
1       5'     4"          17  feet.     Then  multiplying  by  4',  we 

say  4'x4'  =  16"=:l'  and  4";   set  down 

18  ft.  9'    4"         the  4"  and  carry  the  I7,  and  say  4'  times 

4  are  16'  and  1'  is  17'  =  1   ft.    and  5', 

which  being  set  down,  and  the  two  products  added,  we  obtain 
18  ft.  9'  4"  as  the  required  product. 

2.  3.  4. 

ft.     in.  ft.     in.  ft.    in. 

Mult.    8     6'  Mult.    9     10'  Mult.    3     8' 

by    4     3'  76'  7     6' 


34     0'  68     10'  27     6' 

2     1'     6"  4     11'     0" 


Prod.  36     1'  6"            Prod.  73       9'     0" 

5.  6.  7.  8. 

ft-     in.  ft.      in.  ft.    in.  ft.    in. 

Mult.  73'  311'  46'  9     7' 

47'  95'  58'  36' 


33     2'     9"       36     10'    7"        25     6'  33     6'  6'" 


238  MULTIPLICATION  OF  DUODECIMALS. 

9.  10. 

ft.    in.  ft.  in. 

6     4'     3"  3  9'  11" 

4     6'     4"  4  2'  3" 


28     9'     2''     11'"  16     0'       3''     3"'     9"" 

11.  How  many  square  feet  does  a  board  28  ft.   10'  6"  long, 
and  3  ft.  2'  4"  wide  contain?     Ans.  92ft.  2'  10"  6'". 

12.  In  a  board  16  ft.  9'  long,  and  2  feet  3  broad,  how  many 
square  feet  ?     Ans.  37  ft.  8'  3". 

13.  There  is  a  wall  82  ft.   6  in.  high,  and  13  ft.  3  in.  wide. 
How  many  square  feet  does  it  contain?     Ans.  1093  ft.  1'  6". 

14.  There  is  a  room  20  feet  square  and  7  ft.  6  in.  high,  to 
be  plastered  at   lOd.   New  York  currency,   per  square  yard. 
How  many  dollars  will  it  cost  ?     Ans.  $6.94.+ 

15.  There  is  a  yard  58ft.  6  in.  in  length,  and  54ft.  9  in.  in 
breadth.     How  many  dollars  will  it  cost  to  pave  it,  at  5  d.  New 
York  currency,  per  square  yard?    .Arcs.  $18.53. 

16.  If  a  floor  be  59  feet  9  inches  long,  and  24  feet  6  inches 
broad,  how  many  square  yards  does  it  contain  ?     Ans.    162 
yards,  5ft.  10'6;/. 

Note  2d. — If  three  dimensions,  viz.  length,  breadth,  and 
depth  be  given,  the  solid  content  is  found  by  multiplying  them 
successively  into  each  other. 

17.  There  is  a  pile  of  wood  12  feet  6  inches  long,  4  feet 
high,  and  8  feet  6  inches  wide.     How  many  cords  does  ii 
contain  ?     Ans.  3  cords,  41  feet. 

To  reduce  solid  feet  to  cords,  divide  by  128,  that  being  the 
number  of  solid  feet  in  one  cord.  The  required  dimensions 
Of  the  cord,  are  8  feet  long,  4  feet  wide,  and  4  feet  high ; 
since  8x4x4=128. 

18.  How  many  solid  feet  are  there  in   a    block  6  feet  8 
inches  in  length,  4  feet  6  inches  in  height,  and  3  feet  4 
inches  in  width  ?     Ans.  100  feet. 

19.  There  is  a  certain  pile  of  wood  measuring  24  feet  in 
length,  16  feet  9  inches  in  depth,  and  12  feet  6  inches  in 
width.     How  many  cords  are  there  ;  and  how  many  solid  feet 
may  be  daily  consumed  to  have  it  last  one  year  ?     Ans.  39 
cords,  33  feet;  daily  allowance,  13  J  feet,  nearly. 


INVOLUTION.  239 

20.  How  many  square  feet  are  there  in  a  board,  which  mea- 
sures 16  feet,  9  inches  in  length,  and  2  feet,  3  inches  in  breadth? 
Ans.  37  feet,  8 '  3  ". 

QUESTIONS. — What  are  Duodecimals  ?  How  is  the  foot  divided  ? 
"What  is  each  part  called  1  How  is  the  inch  divided  1  What  is  each 
part  called,  &c.  1  When  is  the  foot  divided  duodecimally'?  Repeat 
the  table  of  denominations.  How  may  duodecimals  be  added  1  What 
difficulty  will  be  encountered  in  Multiplication  of  Duodecimals  ? 
Give  an  illustration  of  the  difficulty.  Of  what  denomination  will 
the  product  of  any  two  denominations  be  1  What  is  the  rule  1  What 
is  Note  1st  1  What  is  Note  2d  1 


INVOLUTION. 


A  number  is  involved  by  being  multiplied  into  itself. 

The  number  thus  multiplied  by  itself  is  called  the  root. 

A  power  of  any  number  is  the  product  obtained  by  multiply- 
ing that  number  into  itself.  The  particular  power  produced 
depends  on  the  number  of  successive  multiplications  ;  the  given 
number  always  being  the  first  power,  and  also  the  root  of  the 
succeeding  higher  powers.  The  first  multiplication  then  pro- 
duces the  second  power  ;  the  second  multiplication,  the  third 
power ;  the  third  multiplication,  the  fourth  power,  &c.,  the 
power  obtained  being  always  one  in  advance  of  the  number  of 
multiplications. 

Illustration  :  2= the  first  power,  and  is  also  the  root  of  the 
succeeding  higher  powers. 

2x2=  4,  the  2d  power,  or  square  of  2. 
2x2x2=   8,  the  3d  power,  or  cube  of  2. 
2  X  2  X  2  X  2  =  16,  the  4th  power,  or  biquadrate  of  2. 
2x2x2x2x2  =  32,  the  5th  power  of  2,  &c. 

The  power  to  which  a  number  is  to  be  raised,  is  frequently 
expressed  by  a  small  figure,  called  the  index  of  the  required 
power,  placed  on  the  right  of  that  number,  thus :  42  denotes 
the  second  power  of  4  =  16 ;  and  43  denotes  the  third  power 
of  4=64  ;  and  9s  denotes  the  fifth  power  of  9=59049,  &c. 


240  INVOLUTION. 

A  fraction  is  involved  by  multiplying  the  numerator  and 
the  denominator  each  into  itself,  the  required  number  of  times  ; 
thus  the  square  of  f  is  f  X  f =^5-  The  square  of  %  is  f  £  ;  and 
the  cube  of  the  same  is  J^f ,  &c. 

If  the  given  quantity  be  a  mixed  number,  it  should  first  be 
reduced  to  an  improper  fraction,  before  being  involved ;  thus 
the  second  power  of  2£  is  2s=f,  and  f  xf  =  2T5=6£.  Or, 
proper  and  improper  fractions  may  both  be  reduced  to  deci- 
mals, and  then  involved. 

If  any  number  be  raised  to  two  different  powers,  the  power 
which  is  obtained  by  multiplying  these  two  powers  together, 
is  expressed  by  adding  their  indices,  thus  :  22x23=25=32  ; 
for  22=4,  and  23=8,  and  8x4  =  32;  and  2x2x2x2x2  = 
32-.  Or,  3:'x35=38=6561,for33=27,  and35=243;  and  243 
X  27.3*6561. 

Or,  again,  any  power  of  a  given  number  is  divided  by  another 
power  of  the  same  number  by  subtracting  the  index  of  the  di- 
visor from  the  index  of  the  dividend,  thus  :  25-f-23=22,  for  25 
=  32,  and  23=8,  and  32-^8  =  4,  the  second  power  of  2.  Or 
34-f-32=32,  for  34=81,  and  32=9,  and  81 -=-9  =  9,  the  second 
power  of  3. 

Ex.  1 .  What  are  the  square,  cube,  and  biquadrate  of  3  ?  Ans. 
3x3  =  9,  the  square  ;  3  X  3  X  3 =27,  the  cube  ;  and  3x3x3 
X  3  =  8 1 ,  the  biquadrate . 

2.  What  are  the  square,  cube,  and  biquadrate  of  5  ?     Ans. 
25,  125,  and  625. 

3.  What  are  the  cube  and  biquadrate  of  12?     Ans.  1728 
and  20736. 

4.  Multiply  the  second   and  third  powers    of  4  together. 
What  is  the  product,  and  what  power  of  4  is  it  ?     Ans.  The 
product,  1024,  or  fifth  power. 

5.  What  power  of  3  is  obtained  by  multiplying  its   third 
power  and  its  fourth  power  together  ;  and  what  is  the  number  ? 
Ans.  7th  power,  or  2187. 

Note. — When  the  number  to  be  raised  to  some  given  power 
consists  of  whole  numbers  and  decimals,  the  number  «of  deci- 
mals to  l)e  cut  off  in  the  required  power  is  ascertained  by 
multiplying  the  number  of  decimals  in  the  given  number  by 
the  index  of  the  required  power. 

6.  What  is  the  square  of  26.13?    26.13x26.13  =  6827769. 
Now  to  determine  how  many  decimals  are  to  be  cut  off,  we 


EVOLUTION.  241 

first  notice  that  the  number  of  decimals  in  the  given  number 
is  two,  and  also,  that  the  index  of  the  required  power  is  2, 
therefore,  2x2=4,  the  number  of  decimals  to  be  cut  off. 
Therefore,  682.7769  is  the  required  power. 

7.  What  is  the  cube  of  25.4  ?     Ans.  16387.064. 

8.  Divide  26  by  23,  and  what  power  of  2  will  be  obtained  ? 
Ans.  8,  the  cube  of  2. 


EVOLUTION. 

Evolution  is  the  reverse  of  Involution.  In  Involution  we 
have  the  root  given  to  find  some  required  power ;  but  in  Evo- 
lution a  power  is  given,  and  a  root  required. 

The  relation  between  roots  and  powers  requires  to  be  clearly 
understood. 

A  root  of  a  number  is  obtained  whenever  that  number  is  re- 
solved into  several  equal  factors  ;  and  a  power  of  a  number  is 
obtained  whenever  that  number  taken  as  a  root,  is  multiplied 
into  itself  once  or  more.  Thus,  2  is  the  cube  root  of  8,  be- 
cause it  may  be  resolved  into  three  2's  ;  or  because  2  raised 
to  its  third  power  equals  8,  for  2  x2  X2  =  8.  Also,  8  is  the 
square  root  of  64,  because  the  second  power  of  8  is  64,  or  8  x 
8  =  64.  Again,  9  is  the  square  of  3,  and  3  is  the  square  root 
of  9  ;  27  is  the  cube  of  3,  and  3  is  the  cube  root  of  27.  Roots 
and  powers  are  therefore  correlative  terms. 

The  exact  root  of  some  numbers  cannot  be  obtained.  Such 
numbers  are  called  irrational  powers,  and  their  roots  are  called 
surds.  Thus,  no  root  can  be  obtained,  which,  when  multiplied 
into  itself,  will  produce  2  ;  2  is  therefore  an  irrational  power, 
and  its  root  is  a  surd.  But  a  number  whose  root  can  be  ex- 
actly extracted  is  a  perfect  or  complete  power,  and  its  root  is 
called  a  rational  number.  Thus,  16  is  a  complete  power,  for 
4  is  its  exact  root ;  4  therefore  is  a  rational  number. 

There  are  two  methods  of  expressing  roots.  The  first  and 
more  common  method  is,  by  using  the  character  called  the 
radical  sign ;  written  thus,  V.  This  sign,  without  any  accom- 
21 


242  EVOLUTION. 

panying  index,  always  indicates  the  square  root.  If  other  roots 
are  required,  the  same  radical  sign  is  used,  with  an  index  of 
the  required  root.  Thus,  V9,  is  an  expression  for  the  square 

root ;  v9,  for  the  cube  root ;  y9,  for  the  fourth  root  of  9,  &c. 
V64,  equals  8,  because  8x8=64;  and  v64=4,  because  4 

X4x4=64,  or  v64=2,  for  2x2x2x2x2x2=64,  &c. 
Hence  the  root  is  to  be  taken  as  a  factor  in  producing  its  cor- 
responding power,  as  many  times  as  there  are  units  in  the  in- 
dex of  the  required  root.  The  other  mode  of  expressing  roots 

is  by  means  of  fractional  exponents.     Thus,  65  expresses  the 

square  root;  6  ,  the  cube  root,  and  64,  the  biquadrate  or 
fourth  root  of  6.  The  chief  advantage  of  this  mode  arises 
from  the  fact,  that  not  only  roots  of  numbers  may  be  expressed 
by  it,  but  also  any  required  power  of  a  given  root.  The  de- 
nominator of  a  fractional  index  always  denotes  a  root  of  the 
quantity  to  which  it  is  applied,  while  the  numerator  ex- 

3 

presses  some  power  of  that  root;  thus,  9'  implies  that  the  fourth 
root  of  9  is  to  be  extracted,  and  that  root  raised  to  its  third 

power.  Again,  64s"  implies  that  the  sixth  root  of  64  is  to  be 
extracted,  and  the  root  then  raised  to  its  fifth  power.  But  the 

sixth  root  of  64  is  2,  and  the  fifth  power  of  2  is  32  ;  therefore, 

j> 

646  =32.  Or  a  power  higher  than  the  given  root  may  be  ex- 
pressed ;  thus,  1 6"2,  implies  the  third  power  of  the  square  root 
of  16  ;  but  the  square  root  of  16  is  4,  and  the  third  power  of 

4  is  64  ;  therefore,  16^= 64. 

When  several  numbers  are  to  be  added  and  the  root  of  the 
sum  obtained,  it  may  be  expressed  thus  :  Vf,5-j_i6 ;  which 
implies  that  the  root  of  the  sum  of  65  and  16  is  to  be  obtained. 
If  the  vinculum  over  the  two  numbers  be  rejected,  the  expres- 
sion would  imply,  that  16  is  to  be  added  to  the  square  root  of 
65.  As  the  expression  now  stands,  its  value  is  65+16  =  81, 
and  V8i  =  9.  Or  the  root  of  the  difference  of  two  quantities 
may  be  expressed  in  like  manner,  by  placing  the  minus  sign 
between  them ;  thus,  VcjoUjje,  the  value  of  which  is  90 — 26 
=64,  and  V64  =  8.  Without  the  vinculum  over  the  two 
quantities,  the  expression  would  imply  that  26  is  to  be  taken 
from  the  square  root  of  90. 

The  root  of  the  product  of  several  numbers  is  equal  to  the 


EXTRACTION  OF  THE  SQUARE  ROOT. 


243 


product  of  their  roots.  As  an  illustration,  take  9  and  16.  Their 
product  is  144,  of  which  the  square  root  is  12  ;  that  is,  the 
root  of  their  product  is  12.  The  product  of  their  roots  is  the 
same,  for  the  square  of  9  is  3,  and  of  16  is  4  ;  and  4  X  3  is  12. 
The  same  is  true  of  the  cube  roots,  or  of  any  roots  whatever. 

Take  the  numbers  8  and  27.    y8=2,  y27=3,  and  3  x2=6, 

the  product  of  their  roots.  Again,  27x8=216,  and  V216  =  6, 
the  root  of  their  products. 


EXTRACTION  OF  THE  SQUARE  ROOT. 

The  Square  Root  of  any  number,  is  that  number  which  be- 
ing multiplied  into  itself  once,  will  produce  the  number  given. 

The  following  table  exhibits  the  square  of  all  numbers  from 
1  to  12  : 


Roots, 

M  |2| 

3|    4| 

5| 

6 

!   7 

1    « 

9 

10  | 

11 

1    12 

Squares, 

1  !  1  4  1 

9|  16  | 

25  | 

36 

|4<J 

|64 

81 

100  | 

121 

|  144 

A  square  is  a  figure  bounded  by  four  equal  sides,  and  has 
all  its  angles  right  angles,  or  angles  of  90  degrees.  This  may 
be  seen  in  figure  1st.  Now  the 
area  of  a  square,  that  is,  the  number 
of  square  feet,  or  rods,  &c.,  it  con- 
tains, is  found  by  multiplying  the 
length  of  any  two  sides  together  ; 
or  since  the  sides  are  all  equal,  by 
multiplying  the  length  of  any  one 
side  into  itself ;  or  by  squaring  it. 
As  each  of  the  sides  of  the  annexed 
figure  is  8  feet  in  length,  it  is  obvi- 
ous they  may  each  be  divided  into 
8  equal  parts,  each  of  which  will  be  one  foot  in  length.  Let 
each  side  be  thus  divided  and  the  points  of  division  united,  as 


FIG.  1st. 
8ft. 

90°                         90° 

d 

00 

90°                         90° 

8ft. 


aioL  JUUL  eduu.       i  nu  wiiui        e  33|34|35|3t>|37|38  39,40 \e 

fi^re  is    therefore    divided  into    /41|42|43|44|45|46|47|48/ 
64  equal  parts,  each  of  which  is    /-Ll»       ,  '    .  ,,  ,.K 


just  one  foci  square.      But  8x8     ^49|50|51|5a|53|5. 

ft  nm  c  .1  f  I   ^TI^Ql^OlAnl^l  !£OlCO  1C /<  !  I. 


244  EXTRACTION  OF  THE  SQUARE  ROOT. 

seen  in  figure  2d.     Now  since  PIG.  3d. 

each  side  is  8  feet  long,  the  di- 
visions aa,  bb,  cc,  dd,  &c.  must 
be  just  one  foot  each ;  the  di- 
visions made  by  the  lines  run- 
ning at  right  angles  to  these,  are  d  25|26|27|28|29|30 
also  one  foot  each.  The  whole 


2|   3|  4|  5[  6|  7|  8> 


9|10|U|12|13|14|15|16|6 

l"7|18|19]20|21|22|23|24 


=  64.     Therefore,  the  area  of  a     A57|58|59|60|61|62|63|64JA 
square  is   obtained  by  multiply- 
ing the  length  of  the  side  by  itself;  or,  in  other  words,  by 
squaring  it. 

Now  to  apply  the  above  remarks  to  Evolution,  suppose  we 
have  an  area  equal  to  what  is  given  above,  but  placed  in  a  dif- 
ferent form.  Suppose  it  to  consist  of  a  board  one  foot  wide 
and  64  fe.et  long ;  and  that  it  is  required  to  determine  how  large 
a  square  floor  it  will  exactly  cover.  V64  =  8  feet,  is  the  length 
of  the  side  of  the  required  floor.  Hence,  the  area  of  a  square 
is  found  by  squaring  one  of  its  sides  ;  and  the  length  of  its 
sides  is  found  by  extracting  the  square  root  of  the  given  area. 

The  following  is  the  rule  for  extracting  the  square  root  : 

RULE  1st. — Separate  the  given  number  into  periods  of  two 
figures  each,  by  placing  a  point  or  dot  over  the  place  of  units, 
another  over  the  place  of  hundreds,  and  another  over  the  place  of 
tens  of  thousands,  <fyc. 

2d.  Find  by  trial  the  greatest  square  root  of  the  left  hand 
period,  and  place  it  for  the  first  figure  of  the  root,  after  the  man- 
ner of  the  quotient  in  division. 

3d.  Subtract  the  square  of  this  root  from  the  left  hand  pe- 
riod, and  to  the  remainder  bring  down  the  next  period  of  two 
figures  for  a  dividend. 

4th.  Double  the  root  figure  already  obtained,  for  a  divisor ; 
then,  omitting  the  right  hand  figure  of  the  dividend,  divide  as  in 
Simple  Division,  and  place  the  figure  obtained,  as  the  second 
figure  of  the  root  or  quotient ;  and  also  on  the  right  hand  of 
-the  divisor. 

5th.  Multiply  the  divisor  thus  increased,  by  the  figure  last 
placed  in  the  root,  and  place  the  product  under  the  dividend,  as  in 
Division ;  then  subtract,  and  to  the  remainder  bring  down  the 
next  period  of  two  figures. 


EXTRACTION  OF  THE  SQUARE  ROOT.          245 

6th.  Double  the  root  already  found  for  a  new  divisor,  with 
which  divide  as  before.  Continue  the  operation  till  the  periods 
are  all  brought  down :  the  number  obtained  will  be  the  root  re- 
quired. 

Note  1st. — If  the  number,  the  root  of  which  is  required, 
consist  in  part  of  a  decimal,  place  the  first  point  over  the  unit 
figure,  as  already  directed,  and  point  off  both  ways  from  that 
figure,  allowing  two  figures  to  each  point.  If  the  decimal 
consist  of  an  odd  number  of  figures,  annex  one  cypher  to 
complete  the  last  period. 

2d.  If  after  all  the  periods  have  been  brought  down,  there 
be  a  remainder,  periods  of  two  cyphers  each  may  be  annexed 
and  the  operation  continued.  The  root  figures  obtained  by 
thus  annexing  cyphers  will  be  decimals,  and  must  be  so 
marked. 

3d.  The  number  of  dots  employed  in  pointing  off  the  given 
number,  always  determines  the  number  of  figures  in  the  re- 
quired root. 

Ex.  1.  What  is  the  square  root,  of  1296  ? 

1296  pointed  according  to  the  rule  is  1296.  Hence  we 
know  that  the  root  will  consist  of  two  figures.  The  next  step 
is  to  determine  the  root  of  12,  the  left  hand  period.  This  is 
done  by  trial.  If  4  be  taken,  it  will  be  found  too  large,  since  4 
X  4  =  16.  We  will,  therefore,  take  3.  3x3  —  9,  and  since  9 
is  less  than  12,  it  is  not  too  large,  and  yet  it  is  the  greatest  in- 
tegral quantity,  whose  square  is  less  than  12,  and  is  therefore 
the  root  we  want. 

OPERATION  CONTINUED. 

1296(3 
Root  squared =9 


396  remainder  increased  by  the  next 
period. 

Now  since  we  are  to  have  another  figure  in  the  root,  the  3 
already  obtained  is  3  tens  or  30,  the  square  of  which  is  30  X 
30  =  900  ;  the  12  is  also  1200.  After  subtracting  the  9,  there- 
fore, there  remains  3,  or  300,  and  the  next  period  being  brought 
down,  we  obtain  the  number  396. 

The  next  step  is  to  find  a  divisor,  which  by  the  rule  is  3  -f-  3 
=  6 ;  therefore, 

21* 


246          EXTRACTION  OF  THE  SQUARE  ROOT. 

1296(36 


6  )  3 
In  dividing,  the  right  hand  figure,  viz.  6,  is  omitted. 

Again,  1296(36 
9 

66)396 
396 

0  00 

The  last  root  figure  is  placed  on  the  right  of  the  divisor, 
making  it  66,  and  the  whole  is  multiplied  by  the  root  figure, 
that  is,  66x6  =  396.  I  then  subtract  and  nothing  remains. 
Therefore,  36  is  the  root  required. 

The  operation  is  proved  by  squaring  the  root ;  thus,  36  X  36 
=  1296. 

If  the  given  number  had  consisted  of  three  or  four  periods 
instead  of  two,  the  operation  would  have  been  continued  by 
bringing  down  another  period,  and  then  doubling  the  root  36  for 
a  new  divisor 

Explanation. — We  will  suppose  the  1296  in  the  preceding- 
operation  to  be  so  many  feet  of  boards  one  foot  in  breadth ;  and 
that  it  is  required  to  know  how  large  a  floor,  exactly  square, 
they  will  cover. 

As  has  already  been  said,  the  12  in  the  number  1296  is  1200, 
and  the  root  3  is  so  many  tens,  or  30.  This,  therefore,  is  the 
length  in  feet  of  one  side  of  a  square  floor,  which  1200  feet 
of  the  boards  will  cover,  and  leave  a  remainder  of  3  or  300. 

Now  to  find  the  area  of  a  square, 
(see  fig.  3d,)  we  multiply  the  length  FIG.  3d. 

of  any  two  sides  together,  or  square  30  ft. 

the  length  of  one  side.  Therefore,  30 
X  30 = 900.  We  have  then  disposed 
of  900  of  1296  feet  given,  and  there 
remains  396  feet  to  be  so  added  to 
this  figure  as  to  preserve  its  square 
form.  This  is  done  by  making  equal 
additions  upon  any  two  adjacent 
sides ;  and  hence  we  see  the  obvi- 
ous reason  for  doubling  the  root, 
(which  is  the  length  of  one  side  of 


Area  900  square  feet. 


EXTRACTION  OF  THE  SQUARE  ROOT. 


247 


FIG.  4th. 

36  ft. 

1? 

«in 

cr1 
in 

900  sqr.  ft. 

00 

P 

6  ft. 

S>         180  sqr.  ft. 

30ft. 


the  square,)  for  a  divisor.  But  the  root,  figure  3d,  being  doubled, 
gives  6  for  a  divisor,  and  this  is  contained  in  the  remaining 
figures,  396,  6  times,  (the  right  hand  figure,  viz.  6,  being 
omitted,  agreeably  to  the  rule.)  Now  this  figure,  6,  expresses 
the  breadth  of  the  addition  which  the  remaining  feet  of  boards 
are  sufficient  to  make  to  the  original  square,  as  seen  at  fig.  3d. 
This  addition  is  seen  at  fig.  4th. 
This  diagram  is  not  a  perfect 
square  ;  a  corner  remains  to  be 
filled  up.  We  will,  however, 
before  completing  it,  ascertain 
how  many  of  396  feet,  that 
remained  after  the  first  square 
of  900  feet  was  completed,  are 
here  disposed  of.  The  length 
of  each  addition  being  30  feet, 
and  the  breadth  6  feet,  the 
area  is  30x6  =  180  sq.  feet; 
and  the  area  of  both,  conse- 
quently, is  180+180=  360. 
But  396—360=36.  There- 
fore, 36  feet  still  remain  to  be  added.  The  scholar,  by  refer- 
ence to  the  preceding  diagram,  will  perceive  that  the  corner, 
which  yet  remains  to  be  filled,  to  complete  the  square,  is  just 
6  feet  from  corner  to  corner;  therefore,  6  X  6  =  36.  This  ad- 
dition just  disposes  of  the  remaining  feet  of  boards,  and  com- 
pletes the  square,  as  may  be  seen  at  fig.  5th.  The  area  of  the 
original  square,  as  seen  at  fig. 
3d,  and  also  of  the  several  addi- 
tions made  at  figures  4th  and  5th, 
disposes  of  the  whole  of  the  given 
quantity  of  board  ;  for  900+180 
+  180+36  =  1296.  But  why, 
in  dividing,  is  the  right  hand  figure 
of  the  dividend  omitted  ? 

The  scholar  will  remember 
that  the  points  placed  over  any 
number,  determine  the  number  of 
figures  in  its  root.  (See  Note  3d.) 
Before  performing  the  operation, 
we  therefore  know  that  the  re- 
quired root  of  the  preceding  num- 
ber wiU  consist  of  two  figures.  The  3,  or  left  hand  figure  of 


Fia.  5th. 

36ft. 

i 

| 

800  sqr.  ft. 

6  ft. 

36 

sq.ft. 

180  sqr.  ft. 

30ft. 


248  EXTRACTION  OF  THE  SQUARE  ROOT. 

that  root,  is  therefore  3  tens,  or  30.  This  number,  viz.  30,  is 
consequently  the  true  value  of  the  root  already  found,  which, 
if  doubled,  will  give  60  as  the  divisor,  and  not  6,  as  in  the 
sum.  The  true  value  of  the  divisor  is  therefore  ten  times  as 
great  as  represented  in  the  operation,  and  hence  the  dividend 
is  divided  by  10;  that  is,  its  right  hand  figure  is  omitted,  to 
make  its  value  correspond  with  the  apparent  value  of  the 
divisor. 

Another  peculiar  feature  of  the  operation  consists  in  placing 
the  quotient,  or  root  figure,  on  the  right  of  the  divisor,  thereby 
multiplying  it  into  itself.  For  this,  there  must  also  be  a  reason. 
If  the  scholar  will  examine  figure  4th,  he  will  notice  a  vacant 
corner,  which,  as  the  addition  made  to  each  of  the  two  adjacent 
sides,  is  6  feet,  must  be  just  6  feet  square.  By  placing  the 
root  figure  on  the  right  of  the  divisor,  and  multiplying  it  into 
itself,  this  corner  is  filled  up  and  the  square  completed. 

2.  What  is  the  square  root  of  9801  ? 

OPERATION. 

9  8  0  i  (  9  9  root. 
8  1 

189)1701 
1701 

The  9  in  the  divisor  is  the  last  root  figure,  placed  there  by 
the  rule. 

3.  What  is  the  square  root  of  30138.696025  ? 

OPERATION. 

3013  8. 6  96025  (173.  605 
1 

27)201 
1  89 

343)1238 
1029 

3466)20969 
20796 


347205)  1736025 
1  736025 


EXTRACTION  OF  THE  SQUARE  ROOT.  249 

It  will  be  observed  from  this  example,  that  when  a  period 
is  brought  down,  and  the  number  obtained  is  not  sufficient 
to  contain  the  divisor,  a  cypher  is  to  be  placed  in  the  root,  and 
also  on  the  right  hand  of  the  divisor,  and  the  next  period  is 
then  to  be  brought  down.  For  arranging  the  points  in  this  sum, 
see  Note  1st. 

4.  What  is  the  square  root  of  5499025  ?     Ans.  2345. 

5.  What  is  the  square  root  of   1522756  1     Ans.  1234. 

6.  What  is  the  square  root  of  207936  ?     Ans.  456. 

7.  What  is  the  square  root  of   5700.25  ?     Ans.  75.5. 

8.  What  is  the  square  root  of  74770609  ?     Ans.  8647. 

9.  What   is   the    square   root   of    419112517321?       Ans. 
647389. 

10.  What  is  the  square  root  of  10  ?     Ans.  3.  16227.  -f 

1  1  .  What  must  be  the  length  of  each  side  of  a  square  field, 
in  rods,  in  order  that  the  field  may  contain  40  acres  1  Ans. 
80  rods. 

12.  A  certain  regiment  consists  of  6561  men.     How  many 
must  be  placed  in  rank  and  file,  to  form  them  into  a  square  ? 
Ans.  81  men. 

13.  A  company  of  men  spent  6  £.  13s.  4  d.,  which  was  just 
as  many  pence  for  each  man  as  there  were  men  in  the  company. 
How  many  men  were  there,  and  how  many  pence  did  each 
man  spend  ?     Ans.    There  were  40  men,  and  each  man  spent 
40  d. 

14.  If  a  man  were  to  plant  3969  hills  of  corn  in  a  square, 
how  many  rows  must  he  have,  and  how  many  hills  in  a  row  1 
Ans.  63  of  each. 

Note  4th.  —  To  extract  the  square  root  of  a  vulgar  fraction, 
first  reduce  the  fraction  to  its  lowest  term,  and  then  extract  the 
root  both  of  the  numerator  and  denominator. 


15.  What  is  the  square  root  of 

^  __   9_  and  the  square  root  of  9  is  3      . 
128  —  16  and  ......     16  is  4'  Ans' 

16.  What  is  the  square  root  of  y^-?     Ans.  |-. 

17.  What  is  the  square  root  of  jyf  ?     Ans.  £. 

18.  There  is  an  army  containing  9216  men.     How  many 
men  must  be  placed  rank  and  file,  to  form  a  square  which 
shall  contain  every  man  ?     Ans.  96. 

19.  What  is  the  square  root  of  |§f|-  ?     Ans.  f  . 


250 


EXTRACTION  OF  THE  SQUARE  ROOT. 


APPLICATION    OF    SQUARE    ROOT    TO    PARALLELOGRAMS    AND 
TRIANGLES. 

A  parallelogram  is  an  oblong  figure  having  its  opposite  sides 
equal  and  parallel.     The  adjoining  figure  represents  the  paral- 

FIG.  6th. 


lelogram.  If  we  suppose  this  figure  to  be  10  feet  long,  and 
2  feet  broad,  the  area  of  the  whole  figure  is  evidently  10x2 
—20  square  feet.  The  area  of  a  parallelogram  is  therefore 
found  by  multiplying  its  length  into  its  breadth.  Had  the 
length  of  the  above  figure  been  18  feet,  and  its  breadth  as  given 
above,  the  area  would  have  been  18  x  2  =  36  square  feet ;  and 
this  equals  a  square  figure,  the  sides  of  which  are  6  feet  long, 
for  V36  — 6.  A  square,  equal  in  area  to  a  given  parallelogram, 
is  found  by  extracting  the  square  root  of  the  area  of  that  paral- 
lelogram. 

20.  What  is  the  length  of  the  sides  of  a  square,  whose 
area  shall  be   equal  to  the  area  of  a  parallelogram  32  feet  in 
length  and  2  in  breadth  ? 

Operation  :  32x2  =  64,  and  V64  =  8  feet,  the  side  of  the 
required  square. 

21.  There  is  a  parallelogram  27  feet  in  length  and  3  in 
breadth.     Required  the  sides  of  a  square  of  equal  dimensions. 
Ans.  9  feet. 

22.  Required  the  dimensions  of  a  square,  the  area  of  which 
shall  be  equal  to  the  area  of  a  parallelogram  18  feet  in  length 
and  8  in  breadth.     Ans.  12  feet  square. 

Note  5th. — When  the  area  of  a  parallelogram  is  given,  and 
also  the  ratio  of  its  length  and  breadth,  the  sides  may  be  found 
by  dividing  the  area  by  that  ratio,  and  extracting  the  square 
root  of  the  quotient.  The  root  obtained  will  be  the  required 
breadth,  by  which  divide  the  area,  and  the  quotient  will  be  the 
length. 

23.  If  the  area  of  a  parallelogram  be  288  rods,  and  its  length 
be  twice  as  much  as  its  breadth,  what  is  its  length,  and  what 
its  breadth  ? 


EXTRACTION  OF  THE  SQUARE  ROOT.          251 


Operation  :  288  -r-  2  =  144  and,  ViU=  12,  the  required  breadth  ; 
and  288  -4-  12=24,  the  required  length.  The  reason  of  the  above 
operation  is  obvious.  The  length  being  twice  its  breadth,  if 
the  parallelogram  be  divided  in  the  middle,  the  two  parts  will 
be  equal  and  square,  and  the  square  root  of  one  of  these  parts 
will  evidently  be  the  breadth  of  the  parallelogram.  The  same 
reasoning  may  be  applied  to  parallelograms  of  any  dimensions. 

24.  There  is  a  field  containing  30  acres,  lying  in  the  form 
of  a  parallelogram,  of  which  the  length  is  three  times  the 
width.     What  is  its  length,  and  what  is  its  breadth  1     Ans. 
The  length  is  120  rods,  and  the  breadth  40  rods. 

Note  6th.  —  The  side  of  a  square  of  equal  area  with  a  geo- 
metrical figure  of  any  form,  may  be  found  by  extracting  the 
square  root  of  the  area  of  that  figure. 

25.  I  have  an  irregular  piece  of  land  containing  50f  acres, 
which  I  am  desirous  to  exchange  for  an  equal  number  of  acres 
lying  in  a  square  form.     What  must  be  the  length  of  the  sides 
of  that  square  ?     Ans.  90  rods. 

26.  A  certain  triangular  field  contains    10   acres  of  land. 
What  is  the  length  of  the   sides  of  a  square  containing  the 
same  number  of  acres  ?     Ans.  40  rods. 

The  principle  of  the  square  root  may  be  applied  to  find  the 
length  of  the  sides  of  a  right  angled  triangle,  the  measure 
of  either  two  being  given. 

The  square  of  the  hypoteneuse  or  side  opposite  the  right 
angle,  is  always  equal 

to  the  sum  of  the  squares  ^^\  , 

of  the  base  and  perpen-  ^^ 

dicular.     Therefore, 

If  the  base  and  per- 
pendicular be  given,  and 
the  hypoteneuse  requir- 
ed, square  the  given 
sides,  and  extract  the  JBase. 

square  root  of  their  sum. 

If  the  hypoteneuse,  and  one  of  the  legs  of  the  triangle  be  given, 
to  find  the  other  leg,  square  the  hypoteneuse,  and  from  its  square 
subtract  the  square  of  the  given  side.  The  square  root  of  the 
remainder  will  be  the  length  of  the  required  side. 


252          EXTRACTION  OF  THE  SQUARE  ROOT. 

27.  There  is  a  wall  15  feet  high,  and  in  front  of  it  is 
a  pavement  24  feet  wide.     How  long  a  ladder  is  required  to 
reach  from  the  outside  of  the  pavement  to  the  top  of  the  wall? 
The  hypoteneuse  is  required,  therefore,   15xl5=z225;  and 
24x24  =  576;  then, 225 +  576  =  801 ;  and  V801=:28.3,4-  Ans. 

28.  A  certain  tree  is  broken  off  8  feet  from  the  ground,  and, 
resting  on  the  stump,  touches  the  ground  at  the  distance  of  12 
feet.      What  is  the  length  of  the  part  broken  off?      Ans. 
14.42+  feet. 

29.  There  is  a  fort  standing  by  the  side  of  a  river,  24  yards 
high,  and  a  line  36  yards  long  will  just  reach  from  the  top  of 
the  fort  to  the  opposite  side  of  the  river.     What  is  the  width 
of  the  river?     Ans.  26.832  yards. 

30.  Two  ships  sail  from  the  same  port,  one  due  east,  and 
the   other  due  north.     What   is  the  distance  between  them, 
when  one  has  sailed  100  miles,  and  the  other  168  miles? 
Ans*  195. 5+ miles. 

31.  A  man  shot  a  bird  sitting  on  the  top  of  a  steeple   80 
feet  high,  while  standing  at  the  distance  of  60  feet  from  its 
base.     How  far  did  he  shoot?     Ans.   100  feet. 

32.  A  rope   100  feet  long  attached  to  the  top  of  a  steeple, 
touches  the   ground  when  drawn  perfectly  straight,  20  feet 
from  its  base.     How  high  is  the  steeple  ?      Ans.  98  feet, 
nearly. 

33.  Two  boys  were  playing  with  a  kite,  the  line  of  which 
was  520  feet  in  length.    When  the  string  was  all  out>  one  of 
them  standing  directly  under  the  kite,  and  the  other  holding  the 
string,  the  distance  between  them  was  312  feet.     What  was 
the  perpendicular  height  of  the  kite  ?     Ans.  416  feet. 

QUESTIONS. — When  is  a  number  involved  1  What  is  a  root  1  What 
is  a  power  of  any  number  ?  On  what  does  the  particular  power  pro- 
duced depend  ?  How  does  the  power  obtained  compare  with  the  num- 
ber of  multiplications  in  producing  it  1  How  is  a  required  power  ex- 
pressed 1  What  is  the  figure  denoting  the  power  called  ?  How  is  a 
fraction  involved  ?  If  the  given  quantity  be  a  mixed  number,  what 
must  be  done  1  If  a  number  be  raised  to  two  different  powers,  how  is 
the  power  obtained  by  multiplying  these  two  powers  together,  ex- 
pressed 1  Ho\^r  is  any  power  of  a  given  number  divided  by  another 
power  of  the  same  number'?  When  the  number  to  be  raised  to  a 
power  is  in  part  a  decimal,  how  is  the  number  of  decimals  to  be  cut  off 
from  the  required  power  ascertained  7  What  is  evolution  1  What 
is  a  root  of  a  number  7  What  is  a  power  of  any  number?  What 
are  irrational  powers'?  What  are  surds'?  How  many  methods  are 
there  of  expressing  roots  1  What  is  the  first  method  1  What  root  is 
expressed  by  the  radical  sign  without  any  index  or  exponent  1  If 


EXTRACTION  OF  THE  CUBE  ROOT. 


253 


any  other  root  is  to  be  expressed,  how  is  it  done  !  How  many  times  is 
the  root  to  be  taken  as  a  factor  in  producing  its  correspondieg  power  1 
What  is  the  other  mode  of  expressing  roots'?  What  is  an  advantage 
of  this  mode  1  When  a  fractional  index  is  used,  what  does  the  de- 
nominator denote  ?  What  does  the  numerator  ?  When  several  num- 
bers are  to  be  added,  and  the  root  of  the  sum  extracted,  how  is  the  op- 
eration expressed  7  What  does  the  root  of  the  product  of  several 
numbers  equal  1  Give  an  illustration.  What  is  the  square  root  of 
any  number  1  What  is  a  square  1  How  is  the  area  of  a  square  found  1 
Give  the  illustration.  How  is  the  length  of  the  sides  of  a  square 
found  1  What  is  the  rule  for  extracting  the  square  root  1  What  is  Note 
1st  1  Note  2d  1  Note  3d  1  Why  do  we  take  twice  the  root  for  a  divisor  1 
Why  in  dividing  do  we  omit  the  right  hand  figure  of  the  dividend  1 
Why  do  we  place  the  quotient  figure  on  the  right  of  the  divisor  7 
How  is  the  square  root  of  a  vulgar  fraction  extracted  1  The  vulgar 
fraction  may  first  be  reduced  to  a  decimal  and  the  root  of  the  decimal 
extracted,  if  preferred.  What  is  a  parallelogram  1  How  is  its  area 
found  1  How  may  a  square  equal  in  area  to  a  given  parallelogram  be 
found  1  What  is  Note  5th  1  Note  6th  1  To  what  is  the  square  of 
the  hypoteneuse  of  a  right  angled  triangle  equal  1  If  the  base  and 
perpendicular  be  given,  how  may  the  hypoteneuse  be  found  1  If  the 
hypoteneuse  and  one  of  the  legs  of  a  triangle  be  given,  how  may  the 
other  leg  be  found1?  The  base  and  perpendicular  are  called  the  legs 
of  a  triangle.  How  are  the  operations  in  Square  Root  proved  7  Ans. 
By  multiplying  the  root  into  itself. 


EXTRACTION   OF  THE    CUBE   ROOT. 


IS 


bounded  by  six  equal  plane  surfaces,   each  of 
is  a  square ;  that  is,  the   length,  breadth,  and  depth  of  a 


FIG.  1. 


A  CUBE 

which 

cube  are  equal.  (See  fig.  1st.)  The  area  of  each  of  the  six 
equal  surfaces  is  found  by  squaring  the  measure  of  its  side,  as 
has  already  been  explained  in  Square  Root. 

If  the  adjoining  figure  represent 
a  cubic  block  measuring  three  feet 
in  length,  breadth,  and  thickness,  the 
superficial  area  of  each  face  is  3  x 
3=9  square  feet.  Now  if  the  divi- 
sions marked  by  the  dotted  lines  on 
each  face  of  the  block,  were  extend- 
ed through  it,  in  either  direction,  the 
whole  would  be  divided  into  9  parts, 
each  one  footasquare  and  3  feet  long, 
and  susceptible  of  being  divided  each 
into  three,  and  consequently  the  whole,  into  27  blocks,  each  one 
22 


254  EXTRACTION  OF  THE  CUBE  ROOT. 

cubic  foot.  We  have  then  the  following  general  principle. 
The  content  of  a  solid  or  cubic  body  is  found 'by  multiplying  its 
length,  breadth,  and  thickness  into  each  other ;  or,  what  is  the 
same  in  effect,  by  cubing  one  of  these  dimensions. 

Extracting  the  cube  root  is  a  process,  the  reverse  of  the  pre- 
ceding, that  is,  it  is  finding  the  length  of  one  of  the  sides  of  a 
cubic  body,  the  solid  content  of  that  body  being  given ;  or  it  is 
finding  from  a  given  number,  another  number,  whose  cube  or 
third  power  shall  equal  that  number. 

RULE  1st. — Separate  the  given  number  into  periods  of 
three  figures  each,  by  placing  a  point  first  over  the  unit  figure, 
and  advancing  toward  the  left  when  the  number  consists  of  in- 
tegers only ;  but  to  the  right  and  left  both,  when  it  consists  of 
integers  and  decimals,  and  make  the  last  period  of  the  decimal 
complete,  by  annexing  cyphers,  whenever  necessary. 

2d.  Find  by  trial  the  greatest  cube  root  of  the  left  hand  pe- 
riod, and  place  it  as  in  square  root ;  then  subtract  its  cube  from 
the  same  period,  and  bring  down  the  next  period  of  three  figures 
to  the  remainder  for  a  dividend. 

3d.  Square  the  root  figure  and  multiply  its  square  by  3  for 
a  divisor,  and  see  how  many  times  it  is  contained  in  the  divi- 
dend, omitting  the  first  two  right  hand  figures,  and  place  the 
result  as  the  second  figure  in  the  root. 

4th.  Multiply  the  divisor  by  the  last  quotient  figure,  and, 
placing  two  cyphers  on  the  right  of  the  product, place  the  result 
under  the  dividend.  Also  multiply  the  square  of  this  same  last 
quotient  figure  by  the  former  figure  or  figures  of  the  root,  and 
also  by  3  ;  and,  placing  one  cypher  on  the  right  of  the  product, 
write  it  under  the  preceding  product.  Lastly,  under  this,  write 
the  cube  of  the  last  quotient  or  root  figure,  and  make  the  sum 
of  these  three  numbers  a  subtrahend. 

5th.  Subtract  the  subtrahend  from  the  dividend,  and  to  the 
remainder  bring  down  the  next  period  for  a  new  dividend. 

6th.  To  obtain  a  new  divisor  proceed  as  before,  and  thus 
continue  the  operation,  till  all  the  periods  of  the  given  number 
have  been  brought  down. 

Note  1st. — In  obtaining  each  divisor,  square  the  whole  root 
obtained,  and  multiply  that  square  by  3. 

Ex.  1.  What  is  the  cube  root  of  10648  ? 


EXTRACTION  OF  THE  CUBE  ROOT. 


255 


OPERATION. 

10648(22 


23=8 


Art.  3d,  Rule,  2ax3  =  12  (Div.)  12)2  648    (See  Art.  2d.) 

("12x2  +  00=         2400 
Rule,  Art.  4th,  J   22x2x3  +  0=         240 

23=  8 


Subtrahend,  2648 
0000 

In  dividing,  the  two  right  hand  figures  of  the  dividend  are 
omitted.     (See  rule,  Art.  3.)     Proof,  223=  10648. 

Explanation.— -We  will  suppose  the  number  10648,  in 
the  preceding  sum,  to  be  so  many  feet  of  timber,  one  foot 
square,  and  that  it  is  required  to  find  how  large  a  cubic  pile 
they  will  form  ;  that  is,  what  will  be  the  length,  breadth,  and 
depth  of  a  cubic  pile  containing  that  number  of  solid  feet.  To 
make  each  step  as  clear  as  possible,  we  will  repeat  the  prece- 
ding operation. 

The  number  given  when  pointed,  (see  rule,)  is  divided  into 
two  periods,  viz.  10  and  648.  We  therefore  know  that  the  re- 
quired root  will  consist  of  two  figures,  and  by  trial  we  find  the 
root  of  the  first  or  left  hand  period  to  be  2,  that  is,  2  tens,  or 
20.  Hence,  10648(20.  The 
same  as  is  seen  in  the  above 
operation,  excepting  that  a 
cypher  is  placed  on  the  right 
of  the  root  figure  2,  to  give  it 
its  true  value.  This  gives  the 
linear  measure  of  a  cubic 
block  which  the  10(10.000) 
of  the  given  number  will 
make.  Now,  since  20  is  the 
linear  measure  of  the  cube, 
(that  is,  the  direct  and  not 
diagonal  measure  from  corner 
to  corner,)  it  is  also  the  linear  measure  of  each  of  the  six  equal 
square  faces  of  that  cube.  Therefore,  20  X  20=400,  the  area 
of  each  face  ;  and  400x20  —  8000,  the  number  of  cubic  feet 
required  to  make  a  cubic  body,  whose  linear  measure  is  20  ft. 


FIG.  2. 


256 


EXTRACTION  OF  THE  CUBE  ROOT. 


(See  fig  2.)   By  this  step  we  have  then  disposed  of  8000  of  the 
10648  solid  feet.     Hence, 

10648(20 
8000 


2648 

(This  same  effect  is  obviously  produced  in  the  first  solution  of 
this  sum,  by  cubing  the  2,  and  subtracting  it  from  the  left  hand 
period,  10,  and  then  bringing  down  the  next  period,  648,  to  the 
remainder.)  There  now  remains  2648  feet  to  be  so  added  to 
the  block  already  formed,  that  the  whole  shall  be  a  perfect  cube. 
This  is  done  by  making  equal  additions  on  any  three  of  the 
equal  faces  which  lie  contiguous  to  each  other.  The  reason  of 
this  is  obvious.  A  solid  body  has  length,  breadth,  and  thick- 
ness ;  and  in  a  cubic  body,  these  dimensions  are  all  equal,  and  by 
making  the  additions  as  here  directed,  they  are  equally  increased. 

The  scholar  will  now  understand  why  three  times  the  square 
of  the  root  obtained  is  taken  as  a  divisor.  The  square  of  the 
root  is  the  superficial  area  of  one  of  the  sides  or  faces  of  the 
cubic  block,  and  this  multiplied  by  3  gives  the  area  of  three 
faces  or  sides,  which  is  the  number  of  sides  to  which  equal 
additions  are  to  be  made.  Hence,  dividing  the  quantity  to  be 
added  to  the  cube  now  obtained  by  this  area,  determines  the 
thickness  of  the  addition.  This,  in  the  sum  now  under  con- 
sideration, is  2  feet,  as  seen  at  fig.  3.  But  these  additions  are 
evidently  limited  in  size  to  the  original  block  ;  consequently, 
the  corners  E,  E,  E,  (fig.  3,)  remain  to  be  filled  before  a  per- 
fect cube  is  produced. 

Now   to   determine  the  FIG.  3. 

quantity  here  added.    Each  

face  of   the  original  cube  ^/      g00   ffc 

contains  400  square  feet, 
(see  fig.  2,)  and  this  multi- 
plied by  2,  the  depth  of  the 
addition,  gives  800  solid 

feet  as  the  content  of  the  tiuv  /£,  ^  800t 

addition  made  to  each  face ; 
the  whole  addition  there- 
fore is  800  x  3=2400  solid 
feet,  and  2648  —  2400= 
248  solid  feet,  the  quantity 
yet  remaining  to  be  dis- 
posed of. 


1 

/ 

/ 

800  ft. 

K. 

Solid  60nfcnt 

104:00  f& 

V* 

?ft 

EXTRACTION  OF  THE  CUBE  ROOT. 


257 


OPERATION  CONTINUED. 

10648(22 
8 

22x3  =  12div.     -     -     -    12)2648 
Solid  con.  of  3  additions,  12x2  +  00,(Rule,)  2400 

The  reason  for  placing  two  cyphers  on  the  right  of  the  divi- 
sor multiplied  by  the  root  figure,  is  obvious.  By  referring  to 
a  previous  statement  of  this  sum,  it  will  be  seen  that  the  root 
figure  2,  which,  when  squared  and  multiplied  by  3,  forms  the 
divisor,  is  2  tens,  or  20.  Hence,  20  X  20 =400,  and  400  X  3  = 
1200,  which  would  be  the  divisor,  were  the  full  value  of  the  root 
figure  expressed.  This  deficiency  in  the  divisor  is  made  up 
by  omitting  the  two  right  hand  figures  of  the  dividend  in  divi- 
ding, and  by  placing  two  cyphers  on  the  right  of  the  product  of 
the  divisor  multiplied  by  the  root  figure. 

Our  next  step  will  be  to  fill  the  vacant  corners  as   seen  at 
E,E,E,(fig.  3.) 

The  rule,  after  directing  the  preceding  step,  says,  "  Also 
multiply  the  square  of  this  same  last  quotient  figure  by  the 
former  quotient  figure  or  figures,  and  also  by  3,  and  placing  one 
cypher  on  the  right  of  the  product,  write  it  under  the  preceding 
product"  This  operation 
fills  the  corners  here  allu- 
ded to.  For  since  the  last 
quotient  figure  2,  is  the  thick- 
ness of  the  addition  made, 
its  square,  viz.  4,  is  the 
measure  of  the  corner  to  be 
filled,  and  the  preceding 
quotient  figure  is  the  length 
of  each  corner ;  hence,  4  X 
2  +  0  =  80,  the  solid  content 
of  one  corner,  (see  fig.  4.) 
The  cypher  here  is  added 
because  2,  the  preceding 

quotient  figure,  is  2  tens,  or  20.  The  three  corners  therefore 
require  80  x  3=240  feet  to  fill  them.  The  corner,  C,  still  re- 
mains to  be  filled,  and  8  feet  of  timber  also  remain,  for  10648 
—  10640  =  8.  This  vacant  corner,  C,  measures  exactly  2  feet 
22* 


FIG.  4. 


800  ft. 


^ 

2p 

80ft;. 

600  ft: 

80 

Solid  Content 

10610. 

) 

r> 


258 


EXTRACTION  OF  THE  CUBE  ROOT. 


FIG.  ft. 


800. 


/ 

/ 

80. 

8 

000. 

Solid  Content 
±06*8  Feet. 

80 

in  each  of  its  three  dimen- 
sions ;  hence,  23=8,  is  the 
solidity  of  that  corner;  and 
this  exactly  disposes  of  the 
remaining  timber.  The  cube 
completed  is  seen  at  fig.  5. 
The  contents  of  the  differ- 
ent parts  of  the  completed 
block  are,  fig.  2,  8000+  ;  fig. 
3,  2400+  ;  fig.  4,  240+  ; 
fig  5,  8=10648  feet. 

Note     1st. — In    Square 

Root  we  were  directed  to      

point  off  the  given  number 

into  periods  of  two  figures  each.;  and  in  Cube  Root,  the  direc- 
tion is,  to  allow  three  figures  to  each  period.  The  following  is 
the  reason  : — the  square  of  any  number  always  consists  of  twice 
as  many  figures  as  the  number  itself,  or  one  less  than  twice  as 
many.  The  cube  of  any  number  always  consists  of  three 
times  as  many  figures  as  the  number  itself,  or  one  or  two  less 
than  three  times  as  many.  That  is,  in  Square  Root,  the  left 
hand  period  may  consist  of  one  or  two  figures ;  and  in  Cube 
Root,  the  same  period  may  consist  of  one,  two,  or  three  figures. 

Illustration. — 132=  169,  one  less  than  twice  the  number  of 
figures  squared,  and  to  extract  its  root  it  would  be  thus  pointed, 
169.  462=2 1 16,  twice  the  number  of  figures  squared.  133= 
2197,  two  less  than  three  times  the  figures  cubed,  and  the  left 
hand  period  consists  of  one  figure  only.  253=  15625,  one  less 
than  three  times  the  figures  cubed,  and  the  left  hand  period 
consists  of  two  figures.  993= 970299,  three  times  the  number 
of  figures  cubed,  and  the  left  hand  period  consists  of  three 
figures. 

In  the  following  solution,  the  scholar  will  carefully  compare 
each  step  of  the  operation  with  the  rule.  The  first  thing  to  be 
done,  is  to  form  the  periods,  (Art.  1st,  Rule.)  The  first  figure 
of  the  root  is  then  to  be  determined,  its  cube  subtracted,  and 
to  the  remainder,  the  next  period  of  three  figures  to  be  brought 
down.  (Art.  2d,  Rule.)  He  must  then  proceed  to  obtain  a  divi- 
sor, as  directed  by  Art.  3d,  and,  lastly,  to  determine  the  solidity 
of  the  several  additions  made,  and  to  make  the  result  a  sub- 
trahend. (Art.  4th.) 

Ex.  2.  What  is  the  cube  root  of  12812904  ? 


EXTRACTION  OF  THE  CUBE  ROOT. 
OPERATION. 


259 


12812904(234 
-     8 


22x  3  =  12  (Divisor)    -     -     -     12)4812 
12x3  +  00     -     - 
33     - 


-  3600 

-  540 

27 


4 167= Subtrahend. 


232x  3  =  1587  (Divisor)       -  1587)645904  =  New  dividend. 


1587x4+00     - 

42x23x3  +  0     - 

43      - 


634800 
11040 
64 

645904= Subtrahend. 

000000 
Proof,  2343  =  12812904. 

It  is  obvious  from  the  first  division  by  12,  that  the  divisor  is 
not  contained  in  the  dividend,  in  all  instances,  as  many  times 
as  it  would  be  in  simple  division. 

In  dividing  it  must  be  remembered  to  omit  the  first  two  figures 
on  the  left  hand  of  the  dividend. 

3.  What  is  the  cube  root  of  250047  ?     Ans.  63.     Proof,  633 
=250047. 

4.  What  is  the  cube  root  of  970299  1     Ans.  99.  Proof  as 
before. 

5.  What  is  the  cube  root  of  ]  .953125  1     Ans.  1.25. 

6.  What  is  the  cube  root  of  22069810125  ?     Ans.  2805. 

7.  What  is  the  cube  root  of  183250432  ?     Ans.  568. 

8.  What  is  the  cube  root  of  84.027672  ?     Ans.  4.38. 

9.  What  is  the  cube  root  of  6859  ?     Ans.  19. 

10.  What  is  the  cube  root  of  205379  ?     Ans.  59. 

11.  What  is  the  cube  root  of  432081216  ?     Ans.  756. 

1 1 .  There  is  a  cubic  rock  containing  8000  solid  feet.     What 
is  the  distance  from  corner  to  corner  1     Ans.  20. 


260  EXTRACTION  OF  THE  CUBE  ROOT. 

13.  What  is  the  difference  between  half  of  a  solid  foot,  and 
a  solid  half  foot  ?    Ans.  3  solid  half  feet. 

14.  What  is  the  superficial  area   of  one  of  the  faces  of  a 
cubic  block  containing  4096  solid  feet?     Ans.  256  square  feet. 

15.  What  is  the  side  of  a  cubical  mound,  equal  to  one,  144 
feet  long,  108  feet  broad,  and  24  feet  deep?     Ans.  72  feet. 
Multiply  together  the  several  dimensions  of  the  given  mound, 
and  extract  the  cube  root  of  their  product. 

Note  2d. — AH  solid  bodies  are  to  each  as  the  cubes  of  their 
similar  sides  or  diameters. 

16.  If  a  ball  weighing  8  Ib.  be  6  inches  in  diameter,  what 
will  be  the  diameter  of  another  ball  of  the  same  material,  weighing 
fi4  Ib  ?  a 

8  :  64  : :  63 :  1723   vT&B=Ans.  12  inches. 

17.  If  a  ball  6  inches  in  diameter,  weigh  8  Ib.  what  is  the 
weight  of  another  ball  of  the  same  kind,  measuring  12  inches 
in  diameter  ?     Ans.  64  Ib.     63 :  123 : :  8  :  64. 

18.  What  would  be  the  value  of  a  globe  of  silver,  one  foot 
in  diameter,  if  a  globe  of  the  same,  one  inch  in  diameter,  be 
worth  $6?     A ns.  $10368. 

19.  If  a  globe  of  silver,  one  inch  in  diameter,  be  worth  $6, 
what  is  the  diameter  of  another  globe  of  the  same  metal,  worth 
$10368?     Ans.  12  inches. 

20.  How  many  globes  one  foot  in  diameter,  would  be  required 
to  make  one  globe  27  feet  in  diameter  ?     Ans.  19683. 

21.  Suppose  the  diameter  of  the  sun  to  be   110  times   as 
large  as  that  of  the  earth ;  how  many  bodies  like  the  earth  would 
be  required  to  make  one  as  large  as  the  sun  ?     Ans.   1331000. 

22.  If  a  man  dig  a  square  cellar,  that  will  measure  5  feet 
each  way,  in  one  day,  how  long  would  it  take  him  to  dig  one 
measuring  15  feet  each  way  ?     Ans.  27  days. 

QUESTIONS. — What  is  a  cube  1  How  is  the  area  of  each  face  of  a 
cubic  body  found  1  How  is  the  content  of  a  cubic  body  found  1  What 
is  the  extraction  of  the  cube  root  ^  How  is  the  number  whose  root  is  to 
be  extracted,  to  be  pointed  1  Of  which  period  is  the  root  first  found  1 
What  is  then  done  with  this  root  7  How  many  figures  are  to  be  brought 
down  to  what  remains  ?  How  is  a  divisor  found  1  By  what  do  you 
multiply  the  divisor  7  How  many  cyphers  do  you  place  on  the  right 
of  the  product  1  Where  do  you  place  the  product  1  What  far- 
ther is  done  with  the  quotient  or  root  figure  1  What  does  the  sum  of 
all  these  products  form  ?  What  is  the  fifth  step  of  the  rule  1  How  is 
a  second  divisor  obtained  1  What  is  Note  1st?  Can  we  know  of 
how  many  figures  the  root  will  consist  1  Of  what  is  the  root  fi^mv 
the  linear  measure  1  How  much  of  the  whole  timber  is  disposed  of 


ARITHMETICAL  PROGRESSION.  261 

by  subtracting  the  cube  of  the  quotient  figure  from  the  left  hand  period, 
in  the  operation  taken  for  explanation  ?  How  many  feet  remain  to 
be  added  7  To  how  many  sides  of  a  cube  must  equal  additions 
be  made  to 'preserve  .its  cubic  form'?  Why  is  three  times  the 
square  of  the  root  taken  for  a  divisor?  What  is  determined  by 
dividing7?  How  many  solid  feet  are  disposed  of  by  the  first  addition 
made  to  the  three  faces  of  the  cube  1  Explain  how  the  solid  content 
of  the  addition  to  each  face  is  obtained.  Why,  in  multiplying  the 
divisor  by  the  root  figure,  are  two  cyphers  placed  on  the  right  of  the 
product  ?  How  is  the  deficiency  of  the  divisor  made  up  in  dividing  1 
What  operation  as  directed  by  the  rule  fills  up  the  corners  left  vacant  1 
Why  is  the  square  of  the  quotient  figure  multiplied  by  the  previous 
root  figures  and  a- cypher  placed  on  the  right  hand  of  the  product  1 
How  much  of  the  remaining  timber  is  required  to  fill  the  three  vacant 
corners  1  What  is  the  measure  of  the  corner  left  vacant,  after  the 
preceding  additions  were  made  1  How  is  it  filled  1  Why  in  Square 
Root  do  we  divide  the  given  number  into  periods  of  two  figures  each'? 
Why  is  the  given  number  divided  into  periods  of  three  figures  each, 
in  Cube  Root  1  Of  how  many  figures  does  the  square  of  any  number 
consist  1  Of  how  many  does  the  cube  1  How  are  the  roots  proved  1 
Ans.  By  raising  the  root  to  a  power  of  the  same  name  as  the  root  itself. 


ARITHMETICAL    PROGRESSION. 

Any  series  of  numbers  more  than  two,  increasing  or  decreas- 
ing by  a  constant  and  uniform  difference,  is  called  Arithmet- 
ical Progression,  or  Arithmetical  Series. 

The  series  formed  by  a  continual  addition  of  any  number, 
(called  the  common  difference,)  is  called  the  ascending  series. 
Thus,  3,  5,  7,  9, 1 1 ,]  3,  &c.  is  an  ascending  series,  of  which  the 
common  difference  is  2.  The  reverse  of  this  forms  the  de- 
scending series ;  that  is,  a  series  decreasing  by  a  continual 
subtraction  of  the  common  difference.  Thus,  13,  11,  9,  7,  5, 
3,  &c.  is  a  descending  series. 

The  numbers  constituting  the  series  are  called  terms.  The 
first  and  last  term  of  the  series  are  called  extremes ;  all  the 
intervening  terms  are  called  the  means,  and  the  number  con- 
stantly added  or  subtracted,  is  called  the  common  difference. 
When  the  first  term  and  common  difference  are  given,  the 
series  may  be  indefinitely  extended. 


262  ARITHMETICAL  PROGRESSION. 

In  every  arithmetical  progression,  five  particulars  are  to  be 
noticed ;  of  which,  if  any  three  be  given,  the  remaining  two 
may  be  found.  The  five  particulars  are,  the  first  term,  the  last 
term,  the  common  difference,  the  number  of  terms,  and  the  sum 
of  all  the  terms. 

CASE  1st. — THE  FIRST  TERM,  THE  NUMBER  OF  TERMS,  AND  THE 

LAST  TERM  GIVEN,  TO  FIND  THE  COMMON  DIFFERENCE. 

Ex.  1.  The  first  term  of  an  arithmetical  series  is  2  ;  the 
number  of  terms,  13  ;  and  the  last  term,  38.  What  is  the  com- 
mon difference  ? 

The  series  commences  with  2,  as  the  first  term  ;  therefore, 
38 — 2;=  36,  is  the  amount  of  all  the  additions  made  to  this 
number.  But  the  whole  number  of  terms  being  13,  and  one 
of  these  being  given,  12  terms  must  be  formed  by  adding  the 
common  difference.  Therefore,  36-;- 12  =  3,  the  common  dif- 
ference required.  The  whole  series,  therefore,  is,  2,  5,  8, 11, 
14,  17,  20,  23,  26,  29,  32,  35,  38  ;  thirteen  in  number. 

We  have  then  the  following  rule  for  solving  sums  like  the 
preceding : 

RULE. — Divide  the  difference  of  the  extremes  by  the  number' 
of  terms,  less  one.  The  quotient  will  be  the  common  difference. 

2.  A  man  in  feeble  health,   commenced  a  journey  and  trav- 
eled 9  days.     On  the  first  day  he  traveled  only  3  miles,  but 
afterwards   continued  to  gain  each  day  an  equal  number  of 
miles  on  the  journey  of  the  preceding  day,  till  the  last  day, 
on  which  he  traveled  43  miles.     The  daily  increase  is  re- 
quired?    Ans.  5  miles  per  day  ;  that  is,  43 — 3-^-8  =  5. 

3.  I  owe  a  debt  which  by  agreement  I  am  to  pay  at  17  dif- 
ferent periods.  The  first  payment  is  to  be  $20,  and  the  last,  $100. 
Required   the  common  difference  of  the   several  payments. 
Ans.  $5. 

4.  A  man  had   10  sons  whose  ages   differed  alike;    the 
youngest  of  whom  was  2  years  old,  and  the  oldest  29.     What 
was  the  difference  of  their  ages  ?     Ans.  3  years. 

CASE  2d. — THE  FIRST  TERM,  THE  COMMON  DIFFERENCE,  AND 

THE  LAST  TERM  GIVEN,  TO  FIND  THE  NUMBER  OF  TERMS. 

Ex.  1.  If  the  extremes  be  3  and  51,  and  the  common  dif- 
ference 6,  what  is  the  number  of  terms  ? 


ARITHMETICAL  PROGRESSION.  263 

51 — 3=: 48,  the  amount  of  all  the  additions  made  to  the  first 
term,  3  ;  and  since  each  addition  is  6,  48^6  =  8,  the  number 
of  additions  ;  that  is,  the  number  of  terms  formed  by  adding 
the  common  difference  to  the  first  term  ;  therefore,  8+1=9, 
the  whole  number  of  terms,  or  answer  required. 

From  the  above  we  derive  the  following  rule  : 

RULE. — Divide  the  difference  of  the  extremes  by  the  common 
difference,  and  add  one  to  the  quotient. 

2.  A  man  commenced  a  journey,  and  traveled  the  first  day 
only  4  miles  ;  after  which  he  gained  each  day  6  miles  on  the 
journey  of  the  preceding  day,  and  on  the  last  day  he  traveled 
94  miles.     How  many  days  did  he  travel?     Ans.  16. 

Operation  :  94 — 4  =  90,  and  90-f-6+l  =  16. 

3.  A  man  commenced  a  journey  in  great  haste,  and  traveled 
63  miles  the  first  day ;  but  being  unable  to  continue   at  the 
same  rate,  the   second  day  he  traveled  only  59   miles  ;    and 
thence  continued  to  lose  4  miles  per  day,  till  the  last  day  of 
his  journey,  on  which  he  traveled  1  \  miles.     How  many  days 
did  he  travel?     Ans.  14  days. 

4.  A  man  set  out  on  a  journey  for  the  improvement  of  his 
health;  the  first  day  he   traveled  10  miles,  and  the  last,   65 
miles,  making  each  day  an  advance  of  5  miles  on  the  journey 
of  the  preceding  day.     How  many  days  did  he  travel  ?     Ans. 
12  days. 

CASE  3d. — THE  FIRST  TERM,  THE  LAST  TERM,  AND  THE  NUM- 
BER OF  TERMS  GIVEN,  TO  FIND  THE  SUM  OF  ALL  THE  TERMS. 

Ex.  1 .  Bought  30  yards  of  cloth,  paying  20  cents  for  the 
first  yard  and  50  cents  for  the  last.  What  was  the  whole  cost, 
allowing  each  succeeding  yard  to  increase  in  price  by  a  con- 
stant access  ? 

The  price  of  each  succeeding  yard  increases  by  a  constant 
access  ;  therefore,  the  price  of  the  last  is  as  much  above  the 
average  price  as  the  price  of  the  first  is  below  it ;  hence,  one 
half  of  the  price  of  the  first  and  last  yard  is  the  average 
price.  Therefore,  20+50=:70  and  70-^-2  =  35  cents,  average 
price  ;  hence,  30x35  =  $10.50,  whole  cost. 


264  GEOMETRICAL  PROGRESSION. 

We  have  the  following  rule  : 

RULE. — Multiply  half  the  sum  of  the  extremes  by  the  num- 
ber of  terms ;  the  product  will  be  the  answer. 

2.  Paid  4  cents  for  the  first,  and  $1.21  for  the  last  yard  of 
a  piece  of  cloth  containing  86  yards.     What  was  the  whole 
cost?     Ans.  $53.75. 

3.  How  many  strokes  does  a  regular  clock  strike  in  24  hours  ? 
Ans.  156. 

4.  If  a  person  walk  3  miles  the  first,  and  91  miles  the  last 
day  of  his  journey,  how  far   will  he  have  walked,  allowing 
him  to  have  been  on  his  journey  24  days  ?     Ans.  1128  miles. 

CtUESTioNS. — What  is  arithmetical  progression  1  What  is  the  as- 
cending series  7  What  is  the  descending  series  7  What  is  denoted 
by  the  word,  terms  ?  What  terms  are  the  extremes  1  What  are  the 
means'?  What  is  the  number  constantly  added  or  subtracted  called"?  How 
many  particulars  require  to  be  noticed  1  What  are  they'?  What  is 
Case  1st  ?  What  is  the  rule  1  What  is  Case  3d  1  What  is  the  rule  1 
What  is  Case  3d  1  What  is  the  rule  1 


GEOMETRICAL    PROGRESSION. 

Any  series  of  numbers,  increasing  by  a  common  multiplier, 
or  decreasing  by  a  common  divisor,  is  called  Geometrical 
Progression,  or  Geometrical  Series.  The  common  multiplier 
of  the  ascending  series,  and  the  common  divisor  of  the  descend- 
ing scries,  is  called  the  ratio  of  the  series,  or  the  common  ratio. 
Thus,  if  we  take  3  as  a  first  term,  and  multiply  it  continually 
by  2,  as  a  common  ratio,  we  obtain  the  series  3,  6,  12,  24, 
48,  96, 192,  384,  768,  &c.,  in  which  series  eacnterm  is  obtain- 
ed by  multiplying  the  preceding  term  by  2.  We  have  here  an 
ascending  series.  Or  the  series  may  be  768,  384,  192,  96, 
48,  24, 12,  6,  3,  &c.,  in  which,  each  term  is  obtained  by  divi- 
ding the  preceding  term  by  2.  This  forms  a  descending  series. 
The  several  numbers  thus  produced  constitute  the  terms  of  the 
series  ;  the  first  and  last  of  which  are  called  the  extremes;  and 
the  intervening  ones  are  called  the  means. 


GEOMETRICAL  PROGRESSION.  265 

In  geometrical,  as  in  arithmetical  progression,  there  are  five 
things  to  be  considered  ;  of  which,  if  any  three  be  given,  the 
other  two  may  be  found.  The  five  things  are,  the  first  term, 
the  last  term,  the  number  of  terms,  the  sum  of  all  tJie  terms, 
and  the  ratio. 

CASE  1st. — THE  FIRST  TERM,  THE  RATIO,  AND  THE  NUMBER 

OF  TERMS  GIVEN,  TO  FIND  THE  LAST  TERM. 

RULE. — Raise  the  ratio  to  a  power  whose  index  is  one  less 
than  the  number  of  terms,  and  multiply  this  power  by  the  first 
term.  The  product  will  be  the  number  required. 

Ex.  1.  The  first  term  of  a  geometrical  series  is  2,  and  the 
ratio  3.  What  is  the  12th  term  ? 

Since  the  given  term  2,  is  the  first  of  the  required  series  of 
12  terms,  and  since  each  succeeding  term  is  found  by  multi- 
plying the  preceding  one  by  3,  3  is  evidently  to  be  taken  as 
multiplier,  or  factor,  eleven  times  ;  that  is,  any  term  of  a  series 
is  equal  to  the  first  term  multiplied  by  the  ratio  raised  to  a 
power  one  less  than  the  number  of  terms.  Thus,  2  X  3  X  3  X 
3x3x3x3x3x3x3x3x3u=:3x2  =  354294,  Ans-.  or 
twelfth  term. 

2.  A  person  purchased  a  house  having  8  doors,  and  agreed 
to  pay  for  the  whole,  whatever  value  might  be  attached  to  the 
eighth  door,  by  allowing  $4  for  the  first,  $16  for  the  second, 
and  $64  for  the  third  door,  &c.    What  did  his  house  cost  him  ? 
Ans.  47x4r=  $65536. 

3.  A  boy  purchased  12  oranges,  and  agreed  to  pay  1  cent 
for  the  first,  4  cents  for  the  second,  16  cents  for  the  third,  &c. 
What  was  the  value  of  the  twelfth  orange  ?     Ans.  $41943.04. 

4.  A  man  wishing  to  get  his  horse  shod,  agreed  to  allow  3 
cents  for  the  first  nail,  9  cents  for  the  second,  and  27  cents  for 
the  third,  &c.  ;  .and  to  pay  for  the  whole  the  value  of  the  last 
nail,  of  which  Lere  were  32.     What  was  the  cost  of  shoeing 
his  horse  1     Ans.  $18530201888518.41. 

Note  1st. — It  is  obvious  from  what  was  said  of  Involu- 
tion, that,  if  the  ratio  be  raised  to  two  or  three  different  pow- 
ers, whose  indices,  when  added  together,  equal  the  index  of  the 
required  power,  the  product  of  these  several  powers  will  be  the 
power,  or  number  required.  Thus,  in  the  second  example,  the 
23 


266  GEOMETRICAL  PROGRESSION. 

answer  is  obtained  thus  :  43x44x4=$65536,  the  answer  as 
before. 

5.  A  person  sold   20  yards  of  cloth  as  follows  :    for  the 
first  yard  he  received  3  d.,  for  the  second,  9  d.,  for  the  third, 
27  d.,  &c.     What  was  the  cost  of  the  twentieth  yard  ?     Ans. 
1 4528268 £.  6s.  9  d. 

6.  A  man  purchased  12  horses  ;  for  the  first  horse  he  gave 
only  4  cents,  for  the  second,  he  gave  16  cents,  for  the  third, 
64  cents,  &c.,  and  thus  in  a  quadruple  ratio  to  the  last.     What 
did  the  twelfth  horse  cost  him  ?     Ans.  $167772.16. 

CASE  2d. — THE  EXTREMES  AND  RATIO  BEING  GIVEN,  TO  FIND 

THE  SUM  OF  ALL  THE  TERMS. 

RULE. — Divide  the  difference  of  the  extremes  by  the  ratio 
less  1,  and  the  quotient  increased  by  the  greater  extremes,  will  be 
the  sum  of  the  series. 

Ex.  1.  The  extremes  of  a  geometrical  series  are  2  and 
1458.  What  is  the  sum  of  all  the  terms,  the  ratio  being  3  ? 

1458 — 2=1456,  the  difference  of  the  extremes;  and  3 — 1 
=2,  the  ratio  less  1.  Therefore,  1456-^2=728;  and  728+ 
1458=2186,  the  sum  of  all  the  terms,  Ans. 

2.  A  farmer  has  sheep  in  6  different  pastures.     In  the  first 
pasture  there  are  3,  and  in  the  sixth,  729.     If  the  ratio  of  in- 
crease be  3,  how  many  sheep  has  he  in  all  his  pastures  ?  Ans. 
1092. 

3.  There  is  a  cherry  tree  with  10  branches ;  on  the  first 
branch  there  are  only  2  cherries  ;  on  the  tenth  branch  there 
are  524288.     Now  the  ratio  of  increase  from  one  branch  to 
another  being  4,  how  many  cherries  are  there  on  the  tree  ? 
Ans.  699050. 

CASE  3d. — THE  FIRST  TERM,  THE  RATIO,  AND  THE  NUMBER  OF 

TERMS  GIVEN,  TO  FIND  THE  SUM  OF  THE  SERIES. 

RULE. — Find  the  last  term  by  Case  1st,  and  the  sum  of  the 
scries,  or  of  all  the  terms,  by  Case  2d. 

Ex.  1.  A  gentleman  sold  20  yards  of  cloth,  receiving  for 


GEOMETRICAL  PROGRESSION.  267 

the  first  yard,  3  d.  ;  for  the  second  yard,  9  d. ;  and  for  the  third 
yard,  27  d.  How  much  did  he  receive  for  the  whole,  at  that 
rate? 

Case  1st,  319x 3  =  3486784401,  the  last  term.  Case  2d, 
3486784401—3-^2  =  1743392199.+  3486784401=5230176600 
the  sum  of  all  the  terms  in  pence =2 1792402  £.  10s. 

2.  What  Avould  12  horses  cost,  if  4  cents  were  allowed  for 
the  first,  16  cents  for  the  second,  and  64  cents  for  the  third 
horse,  &c.  the  value  thus  increasing  in  a  quadruple  ratio  to  the 
last  or  twelfth  horse  ?     Ans.  $223696.20. 

3.  A  gentleman  gave  his  daughter,  on  the  day  of  her  mar- 
riage, one  dollar,  promising  to  triple  it  on  the  first  day  of  each 
month  in  the  year.     What  was  the  amount  of  her  portion  ? 
Ans.  $265720. 

CASE  4th. — THE  EXTREMES  AND  NUMBER  OF  TERMS  GIVEN,  TO 

FIND  THE  RATIO. 

RULE. — Divide  the  greater  extreme  by  the  less,  and  the  quo- 
tient will  be  that  power  of  the  ratio,  which  is  equal  to  the  num- 
ber of  terms  less  1 .  The  corresponding  root  will,  therefore,  be 
the  ratio. 

Ex.  1.  The  first  term  of  a  geometrical  series  is  2,  and  the 
last  term  354294,  and  the  number  of  terms  12.  What  is  the 
ratio  ? 

354294—2  =  177147,  and  the  eleventh  root  of  this  number 
is  the  ratio  required ;  therefore,  V177147=3,  the  ratio. 

2.  The  first  term  of  a  certain  series  is  4,  and  the  last  65536, 
and  the  number  of  terms  8.  What  was  the  ratio  ?  Ans.  4. 

QUESTIONS.— What  is  Geometrical  Progression  1  What  is  the  ratio 
of  the  series  1  What  are  the  terms  of  the  series  1  What  terms  of 
a  series  are  called  extremes  1  And  what  are  called  means  1  In 
geometrical  progression,  how  many  things  are  to  be  considered  1  How 
many  of  these  must  be  given  to  find  the  others  1  What  are  the  five 
things  given  1  What  is  Case  1st  1  What  is  the  rule  for  Case  1st? 
What  is  Note  1st  1  What  is  Case  2d  ?  What  is  the  rule  1  What  is 
Case  3d?  What  is  the  rule  1  What  is  Case  4th  ?  What  is  the  rule 7 


268  ALLIGATION. 


ALLIGATION. 

Alligation  is  the  method  of  mixing  several  simples  of  differ- 
ent qualities  j  so  as  to  obtain  a  compound  of  a  mean  or  middle 
quality. 

CASE  1st. — WHEN  THE  QUANTITIES  AND  PRICES  OF  SEVERAL 

SIMPLES  ARE  GIVEN,  TO  FIND    THE    MEAN  PRICE  OF  THE  MIX- 
TURE. 

RULE. — Find  the  total  value  of  the  several  kinds  to  be  mixed, 
and  divide  the  amount  of  this  value  by  the  whole  number  of  ar- 
ticles. 

Ex.  1.  A  farmer  mixed  together  8  bushels  of  rye  worth 
$0.50  per  bushel;  12  bushels  of  corn  worth  $0.65  per  bush- 
el ;  and  6  bushels  of  oats  worth  $0.30.  What  was  the  value 
of  one  bushel  of  the  mixture  ? 

8  bushels  of  rye  at  50  ct.  =  $4.00  ;  12  bushels  of  corn  at  65 
ct. = 87.80 ;  and  6  bushels  of  oats  at  30  ct. = $  1 .80.  And  8  + 
12-j- 6 =26  bushels,  and  $4.00 +$7.80 +$1.80= $13.60,  and 
$13.30-^-26  bushels = $0.523,+  price  of  one  bushel  of  the 
mixture. 

2.  A  grocer  mixed  6  Ib.  of  tea  at   $1.20  per  Ib. ;  12  Ib.  at 
$1.60  ;  and  81b.  at  $1.80.     What  was  the  value  of  one  Ib.  of 
the  mixture  1     Ans.  $  1 .569. 

3.  If  15  bushels  of  wheat  worth  $1.40  per  bushel,  be  mixed 
with  12  bushels  of  rye  at  $0.60  per  bushel,  and  ten  bushels  of 
oats  at  $0.35,  what  is  the  value  of  one  bushel  of  the  mixture  1 
Ans.  $0.856.+ 

4.  If  6  Ib.  of  gold,  20  carats  fine,  be  mixed  with  12  Ib.  at  18 
carats  fine,  what  is  the  fineness  of  the  mixture?     Ans.  18 J 
carats  fine. 

5.  If  6  gallons  of  wine  at  $0.67  per  gallon,  7  gallons  at 
$0.80  per  gallon,  and  5  gallons  at  $1.20  per  gallon,  be  mixed 
together,  what  will  be  the  value  of  one  gallon  of  the  mixture  ? 
Ans.  $0.867.+ 


ALLIGATION.  269 

CASE  2d. — THE  PRICES  OF  SEVERAL  COMMODITIES  BEING  GIVEN, 

TO  DETERMINE  HOW  MUCH  OF  EACH  COMMODITY  MUST  BE 
TAKEN,  TO  FORM  A  COMPOUND  OF  A  CERTAIN  PROPOSED  ME- 
DIUM VALUE. 

RULE. — Write  down  the  prices  of  the  several  simples  under 
each  other,  placing  that  price  which  is  least  in  value  uppermost, 
and  the  remaining  prices  in  the  order  of  their  values. 

Connect,  by  a  line,  any  price  less  than  the  given  mean 
price,  with  one  that  is  greater,  and  continue  thus  to  do  till  they 
are  all  connected ;  then  place  the  mean  price  on  the  left,  and  sep- 
arate it  from  the  other  numbers  by  a  perpendicular  line.  Write 
the  difference  between  the  proposed  price  of  the  mixture,  and  the 
price  of  each  simple,  opposite  the  number  or  numbers  with  which 
that  simple  is  connected ;  and,  finally, 

Notice  whether  more  than  one  difference  stands  opposite  any 
one  price ;  if  so,  their  sum  will  express  the  quantity  of  that 
price  to  be  taken ;  but  if  only  one  difference  stands  there,  that 
will  be  the  quantity  required. 

Note. — One  difference  at  least  must  stand  against  each  price. 

Ex.  1.  How  much,  corn  at  48  cents,  barley  at  36  cents,  and 
oats  at  24  cents  per  bushel,  must  be  taken  to  make  a  com- 
pound worth  30  cents  per  bushel  ? 

'24)   \   6+18=24  bushels  at  24  cents. 

36  $    >  6  ....  6 36  cents. 

48      )   6  ....  6 48  cents. 

The  difference  between  30,  the  mean,  and  24,  is  placed  op- 
posite of  both  36  and  48,  as  it  is  connected  with  them  both ; 
and  the  difference  between  30,  the  mean,  and  36,  and  also 
between  30  and  48,  are  both  placed  opposite  24,  because  these 
numbers  are  both  linked  with  24,  and  the  sum  of  their  differ- 
ences determines  the  number  of  bushels  required  of  that  price. 
Of  the  oats,  therefore,  24  bushels  are  required,  and  of  the  corn 
and  barley,  only  6  bushels  of  each. 

2.  I  have  four  kinds  of  sugar  valued  at  8, 12, 15,  and  18  cents 
per  pound.  How  much  of  each  kind  must  be  taken  to  make  a 
mixture  worth  14  cents  per  pound? 

4  number  of  pounds  at  8  cents. 

1 12  cents. 

2 15  cents. 

6 18  cents. 


Mean  price,  30 


14 


270  ALLIGATION. 

3.  A  grocer  mixed  together  three  kinds  of  tea,  valued  at  6, 
9,  and  10  shillings  per  pound,  so  that  the  compound  was  worth 
8  shillings  per  pound.     How  much  of  each  sort  did  he  take  1 
Ans.  3  Ib.  at  6  s.,  2  Ib.  at  9  s.,  and  2  Ib.  at  10s. 

4.  A  merchant  has  three  kinds  of  wine.     For  the  one  kind 
he  charges  3  s.  4  d.,  for  the  second  5  s.,  and  for  the  third  7  s. 
per  gallon.     How  much  of  each  is  required  to  form  a  mixture 
worth  6  s.  per  gallon  ?    Ans.  12  gal.  at  3  s.  6  d.,  12  gal.  at  5  s., 
and  44  gal.  at  7  s. 

5.  How  much  gold  at  16,  19,  21,  and  24  carats  fine,  will  be 
required  to  form  a  compound  of  20  carats  fine  ?     Ans.  4  parts 
of  16*  1  of  19,  1  of  21,  and  4  of  24  carats  fine. 

CASE  3d. — THE  PRICE  OF  EACH  OF  SEVERAL  SIMPLES,  THE 

QUANTITY  OF  ONE  AND  THE  PRICE  OF  THE  COMPOUND  BEING 
GIVEN,  TO  FIND  HOW  MUCH  OF  EACH  OF  THE  OTHER  SIMPLES 
IS  REQUIRED. 

RULE. — Link  the  several  prices  together,  as  in  the  last  case, 
and  find  their  differences ;  then  multiply  the  given  quantity  by 
the  differences  standing  severally  against  the  other  quantities, 
and  divide  the  product  by  the  difference  standing  against  itself. 
Or  say,  as  the  difference  opposite  the  given  quantity  is  to  the 
given  quantity,  so  are  the  other  differences  severally  to  their  re- 
quired quantities. 

Ex.  1.  How  much  barley  at  30  cents,  rye  at  36  cents,  and 
corn  at  48  cents  per  bushel,  must  be  mixed  with  12  bushels  of 
oats  at  18  cents  per  bushel,  so  that  the  compound  may  be  worth 
22  cents  per  bushel  ? 

30          "I      4=  4. 

36      V I     4=  4. 

48  >    >   f      4=  4. 

18  O  J    8+14+26^:48. 

The  price  of  the  given  quantity  is  48  ;  therefore,  48  :  12  :: 
4:1,  the  quantity  required  at  30  cents  per  bushel.  The  re- 
maining statements  and  answers  are  the  same,  since  the  dif- 
ferences are  all  the  same.  Therefore,  one  bushel  at  30  cents, 
one  at  36  cents,  and  one  at  48  cents,  would  be  required  to  be 
mixed  with  12  bushels  at  1 8  cents,  to  form  a  mixture  worth  22 
cents. 

2.  A  grocer  has  three  kinds  of  beer  for  sale,  valued  at  7  s., 


Mean  price,  22 


ALLIGATION.  271 

5s.,  and  3s.  per  gallon,  which  he  proposes  to  mix  with  20 
gallons  of  a  superior  quality,  worth  6  s.  per  gallon,  so  that  the 
mixture  may  be  sold  at  4  s.  per  gallon.  How  much  of  the 
first  three  kinds  must  he  take?  Ans.  120  gal.  at  3  s.,  and 20 
gal.  at  5s.  and  7s. 

3.  How  much  tea  at  80,  60,  and  40  cents  perlb.  must  be 
mixed  with  30  Ib.  at  $1.00  per  Ib.  so  that  the  mixture  may  be 
sold  at  70  cents  per  Ib.  1     Ans.  10  Ib.  at  80  cents  and  60  cents, 
and  30  Ib.  at  40  cents. 

4.  How  much  water  of  no  value  must  be  mixed  with  100 
gal.  of  wine  at  7  s.  6  d.  per  gal.,  to  reduce  the  price  to  6  s.  3  d. 
per  gallon  ?     Ans.  20  gal. 

CASE  4th. — THE  PRICE  OF  THE  SIMPLES  BEING   GIVEN,  AND 

ALSO  THE  COMPOUND  TO  BE  FORMED,  TO  FIND  HOW  MUCH  OF 
EACH  SIMPLE  MUST  BE  TAKEN. 

RULE. — Connect  the  prices  of  the  simples  as  in  the  prece- 
ding cases,  and  find  the  amount  of  the  differences ;  then  say,  as 
the  amount  of  the  differences  is  to  each  of  the  differences  taken 
separately,  so  is  the  whole  compound  to  the  part  required. 

Ex.  1.  A  compound  of  15  gallons,  which  shall  be  worth  8 
shillings  per  gallon,  is  to  be  made  of  three  sorts  of  wine,  valued 
at  5,  7,  and  12  shillings  per  gallon.  How  much  of  each  kind 
will  be  required  ? 


8 


-     ...     4 4 

.---4 4 

12  ....     34-1    ....  4 


12 
Then,  12  :  4 :  :  15  :  5,  Ans.    5  Ib.  of  each  kind  are  required. 

Proof,  5  s.  X  5=25  s.    7  s.  X  5  s.  =  35  s. ;  and  12  s.  X  5  s.= 
60s.;  and  25  +  35  +  60  =  120  s. ;  and  120  —  8  =  15  gallons. 

2.  I  have  four  sorts  of  tea,  of  which  the  first  kind  is  worth 
1  s.  perlb. ;   the  second  kind,  3  s.  ;   the  third,  6  s. ;   and  the 
fourth,  10s.     How  much  of  each  kind  will  be  required  to  make 
a  compound  of  120  Ib.  worth  4  s.  per  Ib  ?     Ans.  60  Ib.  at  1  s. ; 
20  Ib.  at  3  s.  ;  10  Ib.  at  6  s. ;  and  30  Ib.   at  10  s. 

3.  How  much  of  each  of  four  kinds  of  coffee,  worth  8,  12, 
18,  and  22  cents  perlb.  will  be  required  to  make  a  compound 


272  POSITION. 

of  120  Ib.  worth  16  cents  per  Ib.  ?     Ans.  36  Ib.  at  8  cents  ;  12 
Ib.  at  12  cents  ;  24  Ib.  at  18  cents  ;  48  Ib.  at  22  cents. 

4.  A  gold  beater  has  gold  15,  17,18,  and  22   carats  fine, 
of  which  he  wishes  to  make  a  compound  of  40  oz.   20  carats 
fine.     How  much  of  each  kind  must  he  take  1     Ans.  25  oz. 
22  carats  fine  ;  and  5oz.  of  15,  17,  and  18  carats  fine. 

5.  How  much  water  of  no  value,  and  how  much  wine  at  90 
cents  per  gallon,  must  be  taken  to  make  100  gallons,  worth  60 
cents  per  gallon  1     Ans.  33  £  gallons  of  water,  and  63|  gallons 
of  wine. 

QUESTIONS. — What  is  Alligation  1  What  is  Case  1st  1  What  is 
the  rule  ?  What  is  Case  2d  1  What  is  the  rule  ?  What  is  the  note  1 
What  is  Case  3d  1  What  is  the  rule  1  What  is  Case  4th  1  What  is 
the  rule  1 


POSITION. 

Position  is  a  rule  by  which  answers  are  obtained  to  such 
questions  as  cannot  be  solved  by  the  common  direct  rules,  by 
assuming  any  convenient  number  or  numbers,  and  then  working 
according  to  the  nature  of  the  question. 

J      SINGLE  POSITION. 

When  the  question  can  be  solved  by  the  assumption  of  a 
single  number,  the  operation  is  called  Single  Position. 

The  following  sum  will  serve  for  an  illustration. 

Ex.  1.  A  teacher  being  asked  how  many  scholars  he  had, 
replied,  "  If  I  had  once,  one  half,  one  third,  and  one  fourth  as 
many  more  as  I  now  have,  I  should  have  185."  How  many 
had  he  ?  We  will  suppose  the  number  to  be  24.  As  he  first 
doubles  his  number,  24  must  be  doubled.  To  this  amount, 
one  half  his  original  number,  12,  must  also  be  added.  He 
then  increases  his  number  by  one  third  of  his  original 


SINGLE  POSITION.  273 

number,   viz.    8,   and  also   by   one   fourth,   viz.  6.      Whole 
amount,  74. 

Now  it  is  evident  that  we  have  not  supposed  the  right  num- 
ber, otherwise  the  amount  would  have  been  185,  as  given  in 
the  sum.  We  have,  however,  increased  the  number  we  sup- 
posed, viz.  24,  by, the  same  or  similar  additions,  as  the  teacher 
did  the  true  number  of  his  scholars ;  consequently,  74,  the 
number  we  obtained,  must  have  the  same  ratio  to  24,  the  num- 
ber assumed,  as  185  has  to  the  real  number  of  scholars  in  the 
school.  Therefore,  74  :  24  :  :  185  :  the  number  required ;  viz. 
60.  Proof,  60+60+30+20+15r=185.  . 

We  have  then  the  following  rule  : 

RULE. — Take  any  convenient  number  and  proceed  with  it  ac- 
cording to  the  conditions  of  the  question,  and  observe  the  result  ; 
then  say,  as  the  number  thus  obtained  is  to  the  given  number, 
so  is  the  assumed  number  to  the  true  one.  Or  the  numbers  may 
be  canceled  by  arranging  the  terms  as  directed  in  Simple  Pro- 
portion. 

Ex.  2.  A'man  being  asked  how  much  money  he  had,  replied, 
that  |,  £,  J,  i  of  his  money  added,  made  $57.  How  much 
money  had  he  ?  Ans.  $60. 

3.  What  number  is  that,  which,  being  multiplied  by  9  and 
divided  by  4,  the  quotient  will  be  27  ?     Ans.  12. 

4.  A  man  borrowed  a  sum  of  money  on  interest,  which  in 
10  years  amounted  to  $1800  at  6  per  cent. ;  what  was  the  sum? 
Ans.  $1125. 

5.  Two  boys  were  playing  at  marbles.     Says  one  to  the 
other,  ^,  %,  and  i  of  my  marbles  added  together  make  45,  and 
if  you  can  now  tell  how  many  I.  have,  you  may  have  them. 
How  many  had  he  ?     Ans.  60. 

6.  A  boy  wishing  to  try  the  skill  of  his   companions  in 
figures,  said  he  had  a  pile  of  apples,  of  which,  if  he  gave  ^  to  A., 
\  to  B.,  and  \  to  C.,  there  would  remain  28  for  D. ;  and  re- 
quested them  to  tell  him  how  many  there  were  in  all.     What 
was  the  number-?     Ans.  112. 

7.  A  person  being  asked  his  age,  said  that  if  J  of  the  years 
he  had  lived  were  multiplied  by  7,  and  f  of  them  added  to  the 
product,  the  sum  would  be  292.     How  old  was  he  ?     Ans.  60 
years. 


274  DOUBLE    POSITION. 

8.  A.  saves  ^  of  his  income,  but  B.  who  has  the  same  in- 
come, spends  twice  as  fast  as  A.,  and  thereby  contracts  a  debt 
of  $120  annually.     What  is  their  income  ?     Arts.  $360. 

9.  The  sum  of  A.,  B.,  andC's  ages,  is  132  years.     B'sage 
is   1^  the  age   of  A;  and  C's  age  is  twice  as  great  as  B's. 
What  are  their  respective  ages  ?     Ans.  A's  age  is  24  ;  B's,  36  ; 
and  C's,  72  years. 


.  —  What  is  Position  1    What  is  Single  Position  7    What 
is  the  rule  ?-    How  may  the  operations  be  canceled  1 


DOUBLE  POSITION. 

jp| 

By  Double  Position,  we  solve  such  sums  as  require  two  sup- 
positions. 

In  this  rule,  the  numbers  supposed  to  be  the  true  ones  bear 
no  certain  or  definite  proportion  to  the  required  answers. 

RULE. — Assume  any  two  convenient  numbers  and  proceed 
with  each  according  to  the  conditions  of  the  question,  and  com- 
pare the  result  of  each  with  the  sum  or  result  given  in  the  ques- 
tion, and  find  their  differences.  Call  each  difference  an  error. 

Multiply  the  first  assumed  number  by  the  last  error ;  and 
the  last  assumed  number  by  the  first  error. 

If  both  errors  are  too  great  or  too  small,  divide  the  difference 
of  these  products  by  the  difference  of  the  errors,  and  the  quo- 
tient will  be  the  number  sought.  But  if  one  of  the  errors  be 
too  large,  and  the  other  too  small,  divide  the  sum  of  the  products 
by  the  sum  of  the  errors. 

Note. — The  errors  are  said  to  be  too  large  or  too  small, 
when  by  operating  on  each  supposed  number  according  to 
the  nature  of  the  question,  the  number  obtained  is  greater  or 
less  than  the  corresponding  number  in  the  sum. 

Ex,  1.  Three  men  found  a  purse  of  money  containing  $80, 


DOUBLE    POSITION.  275 

which  they  agree  to  divide  in  such  a  manner,  that  A.  shall 
have  $5  more  than  B,  and  that  B.  should  have  $10  more  than 
C.  ^Vhat  was  each  man's  share  of  the  money  ? 

Suppose,  1st,  that  C.  had  -  $15 
then  B.  had  by  the  conditions,  25 
and  A., 30 

$70,  a  sum  of  money  less  than 
that  found;  therefore,  $80  — $70= $10,  1st  error. 

Again,  suppose  that  C.  had  -  $20 
B.  of  course  must  have  had  -  30 
and  A., 35 

$85,  a  sum  of  money  greater 
than  that  found  ;  therefore,  $85  —  $80= $5,  2d  error. 

If  now  the  above  operations  be  compared  with  the  rule  and 
the  note  following,  it  will  be  seen  that  the  first  error  is  too 
small,  and  the  last  one  too  large  ;  therefore,  15,  number  first 
supposed  X  5,  the  last  error =75;  and  20,  the  number  last  sup- 
posed X  10,  the  first  error =200  ;  and  200+75=275,  the  sum 
of  the  products;  and  10+5  =  15,  the  sum  of  errors.  There- 
fore, 275-^15  =  $18.333  +  ,  C's  share  ;  and  $18.333  +  $10  = 
$28.333  +  ,  B's  share;  and  $28.333+$5  =  $33.333,  A's  share. 

2.  Four  individuals  having  $100  to  divide  among  them- 
selves, agree  that  B.  should  have  $4  more  than  A. ;  C.,  $8  more 
than  B. ;  and  D.  twice  as  much  as  C.  What  was  each  man's 
share  ? 

1st,  suppose  A.  had      $6  2d,  suppose  A.  had  $8 

then  B.  had      -     -     -  10  then  B.  had      -     -  12 

C., 18  C., 20 

and  D.,       -     -     -     -  36  and  D.,       -     -     -  40 

$70  $80 

and  100  —  70  =  30,  1st  error.  Hence,  100— 80=20, 2d  e,rrjr. 

Here  both  errors  are  too  small,  therefore,  6  x  20  =  120  ;  and 
8x30=240;  then,  240—120  =  120,  the  difference  of  the' 
products  ;  and  30  — 20  =  10,  the  difference  of  errors.  There- 
fore, 120-10  =  12,  A's  share  ;  12  +  4=16,  B's  share  ;  and  16 
+  8=24,  C's  share  ;  and  24+24=48,  D's  share.  Proof,  12 
+  16+24+48  =  100. 


276  DOUBLE  POSITION. 

3.  Three  men  hired  a  piece  of  Avail  built,  for  which  they 
paid  $500.     Of  this,  A.  paid  a  certain  part ;  B.  paid  $10  more 
than  A.,  and  C.  paid  as  much  as  A.  and  B.  both.     What  did 
each  man  pay  ?     Ans.  A.  paid  $120  ;  B.  $130;  and  C.  $250. 

Sums  like  the  preceding  are  solved  with  ease  by  analysis. 
Since  we  have  the  sum  they  all  paid,  we  know  that  C.  paid 
$250,  because  he  has  paid  as  much  as  the  other  two,  that  is, 
one  half  of  the  whole.  Therefore,  A.  and  B.  together  paid 
$250.  But  B.  paid  $10  more  than  A,  hence,  250  — 10=240, 
twice  the  number  of  dollars  A.  paid,  and  240-^-2  =  120,  A's 
share;  then,  120+10  =  130,  B's  share;  and  120+130=250, 
C's  share. 

4.  Two  persons  lay  out  equal  sums  of  money  in  trade.     A. 
gains  120  «£.,  and  B.  loses  80  £.     A's  money  was  then  treble 
B's.     With  what  sum  did  they  commence?     Ans.  180  jC. 

5.  A  farmer  hired  a  laborer  40  days,  on  condition  that  he 
should  receive  20  cents  for  every  day  he  wrought,  and  forfeit  10 
cents  every  day  he  was  idle.     At  the  expiration  of  the  40  days 
he  received  $5.     How  many  days  did  he  work,  and  how  many 
was  he  idle  ?     Ans.  He  wrought  30  days,  and  was  idle  10  days. 

6.  What  is  the  length  of  a  fish  whose  head  is  10  inches 
long,  his  tail  as  long  as  his  head  and  half  the  length  of  his 
body,  and  his  body  as  long  as  his  head  and  tail  both  ?  Ans. 
80  inches. 

7.  Two  persons,  A.   and  B.,  have  the  same  income.     A 
saves  ^  of  his,  but  B.,  by  spending  $150  per  annum  more  than 
A.,  at  the  end  of  8  years  finds  himself  $400  in  debt.     What 
was  their  income,   and  how  much  did  each  spend  annually  ? 
Ans.    Income,  $400.     A  spends  $300,  and  B.  $450. 

8.  A  man  bequeathed  his  property  to  his  three  sons,  on  the 
following   conditions  ;   viz.    to  A.,  one  half,  wanting  $50  ;  to 
B.,  one  third  ;  and  to  C.,  the  remainder,  which  was  $10  less 
than  B's  share.     How  much  did  each  son  receive,  arid  what 
was  the  whole   estate?     Ans.  A.  received  $130;  B.  $120; 
and  C.  $110.     The  whole  estate  was  $360. 

*>$,.  A  farmer  bought  a  certain  number  of  oxen,  cows,  and 
calves;  for  which  he  paid  130<£.  For  every  ox  he 
paid  7£. ;  for  every  cow,  5  £.  ;  and  for  every  calf,  l£.  10s. 
There  were  two  cows  for  every  ox,  and  three  calves  for  every 
cow.  How  many  were  there  of  each  kind  ?  Ans.  5  oxen, 
10  cows,  and  30  calves. 

10.  A  person  after  spending  $10'more  than  J  of  his  annual 


PROMISCUOUS  EXAMPLES.  277 

income,  had  $35  more  than  £  of  it  remaining.     What  was  his 
income?     Ans.  $150. 

11.  A  person  has  two  horses ;  he  also  has  a  saddle  worth 
10  £.  If  the  saddle  be  placed  on  the  first  horse,  the  horse 
and  saddle  are  worth  twice  as  much  as  the  second  horse  ;  but 
the  value  of  the  second  horse  with  the  saddle  is  13  £.  less  than 
the  value  of  the  first  horse.  How  much  is  each  horse  worth  ? 
Ans.  The  first  is  worth  56  £.,  and  the  second,  33 £. 

QUESTIONS. — What  is  Double  Position  7  What  relation  do  the 
supposed  numbers  bear  to  the  true  ones  1  What  is  the  rule  1  When 
are  the  errors  said  to  be  too  large  or  too  small? 


PROMISCUOUS   EXAMPLES. 

Ex.  1.  If  460  be  multiplied  by  36,  and  the  product  divided 
by  9,  what  will  the  quotient  be  ?     Ans.  1840. 

2.  What  number  is  that  which,  when  increased  by  £  of  itself, 
will  be  126?     ^71*.  72. 

3.  What  number  multiplied  by  |  will  produce  16  ?     Ans. 
21*. 

4.  What  fraction  multiplied  by  15  will  produce  £  ?     Ans.  -^Q. 

5.  What  number  multiplied  by  32  will  produce  2912  ?  Ans. 
91. 

6.  What  number  divided  by  21  will  give  65  as  a  quotient  ? 
Ans.  1365. 

7.  How  many  nails  are  required  to  shoe  27  horses,  each 
shoe  requiring  8  nails  ?     Ans.  864. 

8.  In  the  counter  of  a  merchant  there  are  four  drawers,  in 
each  drawer,  4  divisions,  and  in  each  division,  $23.75.     How 
many  dollars  do  the  four  drawers  contain  ?     Ans.  $380.00. 

9.  Two  men  depart  from  the  same  place  and  travel  the  same 
way ;  one  travels  36  miles  per  day,  and  the  other  42.     What 
will  be  the  distance  between  them  at  the  end  of  the  8th  day, 
and  how  far  will  each  have  traveled  ?     Ans.  48  miles  apart, 
the  one  having  traveled  288,  and  the  other  336  miles. 

24 


278  PROMISCUOUS  EXAMPLES. 

10.  A  person  owning  f  of  a  ship,  sold  f  of  his  share  for 
$474.     What  was  the  value  of  the  whole  ship,  at  the  same 
rate  ?     Ans.    $1264. 

1 1 .  How  many  men  must  be  employed  to  finish  a  piece  of 
work  in  15  days,  which  would  require  5  men  24  days  ?   Ans. 
8  men. 

)  12.  A  person  being  asked  the  time  of  day,  answered,  "The 
time  past  noon  is  equal  to  £  the  time  till  midnight."  What  was 
the  time  ?  Ans.  36  minutes  past  5. 

13.  In  a  certain  school,  ^  the  scholars  learn  to  read  and 
write ;  £  learn  geography ;  J  learn  grammar ;  and  1 6  study 
astronomy.    What  was  the  number  in  the  school  ?     Ans.  128. 

14.  What  is  the  whole  length  of  a  pole,  J  of  which  stands 
in  the  ground,  16  feet  in  the  water,  and  -|  in  the  air?    Ans. 
213  feet,  4  inches. 

15.  There  is  a  room  12  feet  long,  8  feet  wide,  and  7  feet 
high.     How  much  paper,  2  feet  wide,  will  be  required  to  paper 
the  same  ?     Ans.  46  yards,  2  feet. 

16.  My  horse  and  saddle  are  both  worth  36  £.  12  s.,  and 
my  horse  is  worth  7  times   as  much  as  my  saddle.     What  is 
the  value  of  each  ?     Ans.   My  horse  is  worth  32  £.  0  s.  6  d., 
and  my  saddle,  4  £.  11  s.  6  d. 

17.  There  is  a  cistern  having  3  faucets,  the  largest  of  which 
will  empty  it  in  one  hour,  the  second  in  2  hours,  and  the  third 
in  3  hours.     In  what  time  will  they  all  empty  it,  if  opened  at 
the  same  time  1     Ans.  32^-  minutes. 

18.  Divide  1500  acres  of  land  between  A.,  B.,  and  C.,  so 
that  A.  shall  have  150  acres  more  than  B.,  and  B.  100  acres 
more  than  C.     Ans.  A.  has  633£,  and  B.  483$,  and  C.  383|. 

19.  A  certain  pasture  will  feed  324  sheep  7  weeks.     How 
many  must  be  turned  away,  in  order  that  it  may  be  sufficient 
for  the  remainder  9  weeks  ?     Ans.  72. 

20.  A  merchant  beught  120  gallons  of  melasses  for  $45. 
How  must  he  sell  the  same  per  gallon,  t<J  gain  15  per  cent.  ? 
Ans.  0.43.4- 

21.  If  a  family  of  8  persons  consume  $200  worth  of  pro- 
vision in  9  months,  how  much  will  18  persons  consume  in  a 
year  ?     Ans.  $600. 

22.  A  man  left  his  son  a  fortune,  J  of  which  he  spent  in  3 
months  ;  and  in  6  months  n\ore  he  spent  f  of  the  remainder, 
when  he  had  only  $1500  remaining.     What  was  his  fortune  ? 
Ans.  $13500 

23.  A  young  man  received  350  £.  as  his  share  of  his  fa- 


PROMISCUOUS  EXAMPLES.  279 

ther's  estate,  which  was  |-  of  his  elder  brother's  portion  ;  and 
the  elder  brother's  portion  was  3  of  the  whole  estate.  What 
was  the  whole  estate  ?  Ans.  2333  £.  6s.  8  d. 

24.  If  356  persons  consume  75  barrels  of  provision  in  9 
months,  how  many  barrels  will  500  men  consume  in  the  same 
time?     Ans.  102ff  barrels. 

25.  A.  can  mow  an  acre  of  grass  in  5|  hours  ;  B.  can  mow 
2  acres  in  9  hours.     In  what  time  will  they  both  mow  12^ 
acres  1     Ans.  30-Lf-  hours. 

26.  If  6  men  build  a  wall  20  feet  long,  6  feet  high,  and  4 
feet  thick,  in  16  days,  in  what  time  will  24  men  build  one  200 
feet  long,  8  feet  high,  and  6  feet  thick  ?     Ans.  80  days. 

27.  A  farmer  being  asked  how  many  sheep  he  had,  replied, 
that  in  one  pasture  he  had  £  of  his  whole  flock,  in  another  J, 
in  another  £,  and  ^  in  another,  and  that  a  fifth  pasture  con- 
tained 450  sheep.     How  many  did  the  live  pastures  contain  ? 
Ans.  1200. 

28.  If  |  of  a  gallon  of  wine  cost  f  of  a  pound,  New  York 
currency,  what  will  f  of  a  tun  cost  in  dollars  and  cents  ?  Ans. 
$262.50. 

29.  If  f  of  an  ounce  cost  $  of  a  shilling,  how  many  dollars, 
each  8  s.  will  a  pound  cost  ?     Ans.  $2.62^. 

30.  Bought  36  bags  of  rice,  each  weighing  84  lb.,  tret  4  Ib. 
per  104.     What  will  the  whole  neat  weight  amount  to  in  fed- 
eral money,  at  8  d.  New  York  currency,  per  pound  ?     Ans. 
$242.307.+ 

31.  In  a  certain  orchard,  ^  of  the  trees  bear  apples,  £  pears, 
i  plums,  and  60  bear  peaches,  and  40  cherries.     How  many 
trees  are  there  ?     Ans.  1200. 

32.  Sold  goods  to  the  amount  of  $560,  by  which  I  lost  18 
per  cent.,  whereas  I  ought  to  have  gained  12  per  cent.    What 
was  my  real  loss?     Ans.  204. 877.+ 

33.  What  is  that  number  of  beggars  to  whom  if  I  give  3 
pence  apiece  I  shall  want  8  pence  more  than  I  now  have,  but  if  I 
gjjpre  them  2  pence  apiece,  I  shall  have  3  pence  left.     Ans.  11. 

t.  How  much  sugar  at  9  d.  per  lb.  must  be  given  in  ex- 
ge  for  492  lb.  of  rice  at  3  d.  per  lb.  ?     Ans.  164  lb. 

35.  Reduce  i  of  f  of  f  of  ^  of  T3T  of  f|,  to  a  simple  fraction. 
Ans.  r/2- 

36.  What  is  the  premium  on  $1800,  at  15  per  cent.  ?    Ans. 
$270. 

37.  A  father  of  12  children  said  he  was  24  years  old  when 
his  oldest  child  was  born,  and  that  just  a  year  and  a  half  inter- 


280  PROMISCUOUS  EXAMPLES. 

vened  between  the  birth  of  each  two  of  his  children.  What 
was  the  age  of  the  father,  at  the  birth  of  his  youngest  child  ? 
Ans.  40^. 

38.  Bought  16  bales  of  goods  in  London  for  96  £.  ;  paid 
4  £.  for  shipment  to  New  York.     How  ought  I  to  sell  the 
same  in  federal  money  to  gain  20  per  cent.  ?  Ans.  $533.333.+ 

39.  Bought  a  quantity  of  Irish  linen  in  Dublin  for  50  £., 
and  paid  10«£.  for  the  shipment  of  the  same.     How  must  I 
sell  the  whole  in  federal  money  to  gain  30  per  cent.  ?     Ans. 
$319.80. 

40.  What  is  the  interest  on  $462.50  for  3  years,  6  months, 
and  12  days  ?     Ans.  $98.05. 

41.  Suppose  there  to  be  two  silver  cups,  having  one  cover, 
which  weighs  5  oz. ;   and  suppose  them  to  be  such,  that  if 
the  cover  be  placed  on  the  smaller  cup,  the  whole  weighs 
twice  as  much  as  the  greater  cup ;  but  if  it  be  placed  on  the 
greater  cup,  the  whole  weighs  three  times  as  much  as  the 
smaller  cup.     What  is  the  weight  of  each  cup  ?     Ans.  The 
smaller  cup  weighs  3  oz.,  and  the  larger,  4  oz. 

42.  A.  can  do  a  piece  of  work  in  6  days  ;  B.   can  do  the 
same  in  1 1  days.     In  what  time  will  they  both  accomplish  the 
work,  if  they  labor  together?     Ans.  3^-y  days. 

43.  A.  can  do  a  piece  of  work  in  7  days  ;  B.  in  12  days; 
C.  in  6  days  ;  and  D.  in  4  days.     In  what  time  will  they  ac- 
complish the  same  work,  if  they  all  labor  upon  it  at  the  same 
time  ?     Ans.  If  day. 

44.  A  person  being  asked  the  time  of  day,  replied  that  it 
was  between  5  and  6  o'clock,  and  that  the  hour  and  minute 
hand  were  precisely  together.     What  was   the  time  ?     Ans. 
27^-  minutes  past  5  o'clock. 

45.  How  many  square  feet  are  there  in  a  board  16  feet  8 
inches  long,  and  9  inches  broad  ?     Ans.  12|  square  feet. 

46.  The  linear  measure  of  a  cubic  block  being  8  inches, 
how  many  cubic  blocks,  each  one  solid  inch,  does  it  contain  ? 
Ans.  512. 

47.  If  a  person  own  J  of  a  ship,  and  sell  f  of  his  shai 
$1500,  what  is  the  value  of  the  whole  ship,  at  the  same  rfle  ? 
Ans.  $6000. 

48.  Divide  f  of  f  of  T7^  of  if  by  |  of  f  of  |.    Ans. 

49.  In  a  thunder  storm,  I  observed  the  time  between  the 
flash  of  the  lightning  and  the  report  to  be  one  minute  and  a 
half.     How    liir  distant  was  the  lightning,   allowing  sound  to 
travel  1142  feet  in  a  second?     Ans.  19|-J-  miles. 

50.  Bought  84  apples  at  the  rate  of  2  for  a  penny,  and  114 


PROMISCUOUS  EXAMPLES.  281 

at  the  rate  of  3  for  a  penny ;  the  whole  of  which  I  afterwards 
sold  at  the  rate  of  9  for  4  pence.  Did  I  gain  or  lose,  and  how 
much  ?  Ans.  I  gained  8  d. 

51.  A  line  44  yards  in  length  will  just  reach  from  the  top  of 
a  steeple  to  the  opposite  side  of  the  street,  which  is  24  yards 
wide.     How  high  is  the  steeple  ?     Ans.  36.87+  yards. 

52.  A  tree  36  feet  high  stands  by  the  side  of  a  stream  27 
feet  wide.     How  many  feet  from  the  top  of  the  tree  to  the  op- 
posite side  of  the  stream  ?     Ans.  45. 

53.  There  is  a  cistern  having  two  pipes  leading  into  it,  one 
of  which  will  fill  it  in  30  minutes,  and  the  other  in  45  min- 
utes.    In  what  time  will  both,  running  together,  fill  the  same  ? 
Ans.  18  minutes. 

54.  How  many  bricks  9  inches  long  and  4  inches  broad 
will  pave  a  yard  40  feet  square  ?     Ans.  6400. 

55.  A  certain  cistern  has  two  pipes  leading  into  it,  and  one 
leading  out  of  it.     Of  the  two  leading  into  it,  one  will  fill  it  in 
50  minutes,  and  the  other  in  75  minutes  ;  while  the  one  lead- 
ing from  it,  will  empty  it  in  60  minutes.     Now  if  all  three  be 
opened  at  the  same  time,  in  what  time  will  the  cistern  be  filled  ? 
Ans.  1  hour. 

56.  What  is  the  difference  between  6  dozen  dozen,  and  half 
a  dozen  dozen  ?     Ans.  792. 

57.  If  -|  of  i  of  £  of  a  ship  be  worth:^-  of  -J  of  ij  of  her 
cargo,  valued  at  $2400,  what  is  the  value  of  both  ship  and 
cargo?     Ans.  $5415.384.+ 

58.  If  1 1  men  can  build  a  house  in  5  months  by  working 
12  hours  per  day,  in  what  time  will  they  complete  it  if  they 
work  only  8  hours  per  day  ?     Ans.  7-|-  months. 

59.  Bought  a  pipe  of  wine  for  $84,  from  which  12  gallons 
leaked  out.     Now  what  shall  I  gain,  if  I  sell  the  remainder  at 
12^  cents  a  pint  ?     Ans.  $30. 

60.  A.  alone  can  do  a  piece  of  work  in  12  days,  and  in  con- 
nection with  B.,  in  8  days.     In  what  time  can  B.  perform  the 
same  alone  ?     Ans.  24  days. 

61.  If  A.  can  do  a  piece  of  work  in  12  days,  and  B.  in  24 
days,  in  what  time  will  they  both  do  it  1     Ans.  8  days. 

62.  Three  men,  A.,  B.,  and  C.,  can  do  a  piece  of  work  in 
18  days.     A.  alone  can  do  it  in  36  days  ;  B.  in  54  days.     In 
what  time  can  C.  do  it  ?     Ans.  108  days. 

63.  Four  men  can  do  a  piece  of  work  in  15  days.    A.  alone 
can  do  it  in  40  days,  B.  in  60  days,  and  C.  in  80  days.     In 
what  time  will  D.  do  the  work  alone  ?     Ans.  80  days. 

24* 


282  PROMISCUOUS  EXAMPLES. 

64.  A  gentleman  left  his  son  a  fortune,  \  of  which  he  spent 
in  3  months ;  \  of  |-  of  the  remainder  lasted  him  6  months 
longer,  when  he  had  only  $1200  left.     What  was  his  whole 
fortune  ?     Ans.  $6400. 

65.  A  farmer  bought  a  yoke  of  oxen,  a  horse  and  a  cow,  for 
$250.     For  the  oxen  he  paid  twice  as  much  as  for  the  horse, 
and  for  the  horse  three  times  as  much  as  for  the  cow.     What 
did  he  pay  for  each  ?     Ans.  For  the  oxen,  $150 ;  for  the  horse, 
875  ;  and  $25  for  the  cow. 

66.  Three  men  start  at  the   same  time  to  travel  round  an 
Island  80  miles   in  circumference,  and  agree  that  each  shall 
continue  to  travel  at  the  same  rate  till  they  all  come  together 
again,  and  that  the  first  shall  travel   5  miles,  the  second,  6 
miles,  and  the  third,  7  miles  per  day.     In  what  time  will  the 
three  come  together,   and  how  far  will  each  have  traveled  1 
Ans.  80  days.     The  first  will  have  traveled  400  miles ;  the 
second,  480  miles  ;  and  the  third,  560  miles. 

67.  How  will  it  affect  the  distance  traveled  by  each,  pro- 
vided they  travel  5,  7,  and  9  miles  respectively  ?     Ans.  The 
distance  of  the  first  will  be  400  miles  ;  of  the  second,  560  ; 
and  of  the  third,  720. 

68.  Divide  1200  acres  of  land  between  A.,  B.,  and  C.,  so 
that  B.  may  have  75  acres  more  than  A.,  and  C.  95  acres 
more  than  B.     What  will  be  the  share  of  each  one  1     Ans. 
A's  share  will  be  318£;  B's,  393£ ;  and  C's,  488J  acres. 

69.  Three  men  met   at  an  Inn,  two  of  whom  brought  pro- 
vision with  them.     The  third  not  having  brought  any,  pro- 
posed that  they  should  eat  together,  and  then  he  would  pay  his 
proportion.  The  proposal  being  accepted,  A.  produced  5  loaves, 
and  B.  4  loaves,  all  of  which  they  ate  up,  and  C.  as  his  share, 
paid  9  pieces  of  money.     With  this,  A.  and  B.  were  satisfied, 
but  could  not  agree  as  to  the  division.     What  should  each  have 
received?     Ans.  A.  6,  and  B.  3  pieces  of  the  money. 

70.  If  A.  can  reap  a  field  of  grain  in  12  days,  and  B.  in  16 
days,  in  what  time  will  both  do  it,  working  together  ?     Ans. 
6f  days. 

71.  What  number  must  be  added  to  ^  of  4560,  to  make  the 
same  500  1     Ans.  369f . 

72.  If  the  head  of  a  fish  be  9  inches  long,  its  tail  as  long 
as  its  head  and  half  the  length  of  its  body,  and  its  body  as 
long  as  its  head  and  tail  both,  how  long  is  the  fish  ?     Ans. 
72  inches. 

73.  If  6  pairs  of  hose  are  equal  in  value  to  4  pieces  of  Hoi- 


PROMISCUOUS  EXAMPLES.  283 

land,  and  6  pieces  of  Holland  to  14  yards  of  satin,  and  12  yards 
of  satin  to  8  pieces  of  lace,  and  9  pieces  of  lace  to  8  £.  2s., 
how  many  pairs  of  hose  may  be  bought  for  2  £.  16s.?  Ans. 
24  pairs. 

74.  A  man  was  hired  50  days,  on  condition  that  for  every 
day  he  worked,  he  should  receive  6  s.,  and  for  every  day  he 
was  idle,  he  should  forfeit  2  s.     At  the  expiration  of  the  time, 
he  received  $27.50.     How  many  days  did  he  work,  and  how 
many  was  he  idle  ?     Ans.   He  labored  40  days,  and  was  idle 
10  days. 

Note. — The  currency  of  the  preceding  sum  is  that  of  New 
York. 

75.  A  hare  starts  40  yards  in  advance  of  a  greyhound,  and 
is  not  perceived  by  him  till  she  has  been  up  40  seconds.    She 
scuds  away  at  the  rate  of  10  miles  per  hour,  and  the  hound 
pursues  after  her  at  the  rate  of  18  miles  per  hour.     In  what 
time  will  the  houtid  overtake  the  hare,  and  how  far  will  he 
have  run?     Ans.    The  required  time  is   60^  &ec. ;  the  dis- 
tance run,  530  yards. 

76.  If  8  men  can  build  a  wall  15  rods  long  in  10  days,  how 
many  men  will  be  required  to  build  45  rods  of  wall  in  5  days  ? 
Ans.  48  men. 

77.  A  man  when  he  married  was  three  times  as  old  as  his 
wife  ;  after  they  had  been  married  15  years,  his  age  was  only 
double  that  of  his  wife's.     How   old  were   they  when  they 
married  ?     Ans.  The  man  was  45  years,  his  wife  15  years  old. 

78.  There  is  in  a  pasture  a  certain  number  of  sheep,  cows, 
and  oxen  ;  there  are  twice  as  many  sheep  as  cows,  and  three 
times  as  many  cows  as  oxen,  and  the  whole  number  is  80. 
How  many  are  there  of  each  kind  ?     Ans.  8  oxen,  24   cows, 
and  48  sheep. 

79.  Sold  coffee  at  15  cents  per  pound,  and  thereby  lost  10 
per  cent,  on  the  first  cost.     Afterwards  sold  a  quantity  of  the 
same  for  $525,  by  which  I  gained  40  per  cent.     What  was  the 
quantity  sold,  and  what  the  price  per  pound  ?     Ans.  Quantity, 
20  cwt.  1 0  Ib. ;  price,  23£  cents. 

80.  Two  sons,  one  11  and  the  other  16  years  of  age,  re- 
ceived a  bequest  of  $10000,  to  be  so  divided  between  them 
that  the  shares  being  put  on  interest  at  5  per  cent,  should 
amount  to  equal  sums,  when  they  became  respectively  21  years 
of  age.     What  were  the  shares  of  each?     Ans.  The  elder 
received  $5454T°r,  and  the  younger,  $4545^T. 


APPENDIX. 


Prob.  1.  To  find  the  greatest  common  measure  of  two  or 
more  numbers. 

Note  1st. — The  greatest  common  measure  of  two  or  more 
immbers,  is  the  greatest  number  that  will  divide  them  separately 
without  remainders. 

RULE. — If  two  numbers  only  are  given,  divide  the  greater  of 
them  by  the  less,  and  if  nothing  remain,  that  divisor  is  the  com- 
mon measure ;  but  if  there  be  a  remainder,  divide  the  preceding 
divisor  by  it ;  and  so  continue  to  divide  each  preceding  divisor 
by  the  last  remainder,  till  the  division  is  effected  without  re- 
mainder ;  the  last  divisor  will  be  the  common  measure  required. 
When  more  than  two  numbers  are  given,  find  the  common  meas- 
ure of  any  two  of  them  first,  and  then  of  that  common  measure, 
and  either  of  the  remaining  numbers.  This  process  carried 
through  all  the  numbers  will  give  their  greatest  common  measure. 

Ex.  1.  What  is  the  greatest  common  measure  of  72  and 
108? 

OPERATION. 
72)108(1 

72 

3  6)  72(2 
72 

00 

36  is,  therefore,  the  common  measure  required.  Proof,  108 
—  36  =  3,  and  no  remainder.  72  -H  36 =2,  and  no  remainder. 

2.  What  is  the  greatest  common  measure  of  27  and  99  ? 
Ans.  9. 


APPENDIX.  285 

3.  What  is  the  greatest  common  measure  of  25,  45,  and  90  ? 
Ans.  5. 

4.  What  is  the  greatest  common  measure  of  16,  32,  48,  and 
96?     Ans.  16. 

Prob.  2.  To  determine  how  many  different  positions  any 
given  number  of  objects  may  assume  with  regard  to  each  other. 

RULE. — Represent  the  number  of  objects  by  the  figures  1,  2, 
3,  4,  5,  fyc.  making  the  numbers  of  figures  equal  to  the  number 
of  objects.  The  product  of  these  figures  will  determine  the 
number  of  changes. 

Ex.  1.  How  many  changes  may  be  made  by  the  first  three 
letters  of  the  alphabet  ? 

Operation  :  1x2x3  =  6,  Ans.  These  changes  are  as  fol- 
lows :  1st,  a,  b,  c;  2d,  a,  c,  b;  3d,  b,  a,  c;  4th,  b,  c,  a;  5th, 
c,  b,  a;  6th,  c,  a,  b. 

2.  How  many  changes  may  be  made  by  the  first  6  letters  of 
the  alphabet  ?     Ans.  720. 

3.  At  a  certain  boarding  house  there  are  12  boarders.     How 
many  different  positions  may  they  occupy  at  the  table  ?     Ans. 
479001600. 

4.  Five  men  engaged  board  at  a  tavern  for  as  many  days  as 
the  landlord  might  be  able  to  seat  them  in  different  positions. 
How  long  did  they  remain?     Ans.  120  days. 

Prob.  3.  To  find  the  area  of  a  square. 

RULE. — Multiply  the  length  of  one  of  the  sides  by  itself;  or, 
square  its  linear  measure.  (See  fig.  2d,  Square  Root.) 

Ex.  1.  There  is  a  room  just  8  feet  square,  what  is  the  area? 
Operation,  8x8  =  64  square  feet,  Ans. 

2.  What  is  the  area  of  a  floor  19  feet   square  ?     Ans.  361 
square  feet. 

3.  What  is  the  area  in  square  inches  of  a  board  3  feet  square  ? 
Ans.  1296  square  inches. 

Prob.  4.  To  find  the  area  of  a  parallelogram. 

RULE. — Multiply  the  length  of  the  parallelogram  by  its  breadth. 
(See  fig.  6th,  Square  Root.) 


286  APPENDIX. 

Ex.  4.  How  many  square  feet  are  there  in  a  floor  16  feet 
long  and  12  broad  ?  16  x  12  =  192  square  feet,  Ans. 

Note  1st. — If  the  dimensions  of  a  field  in  the  form  of  a  square 
or  parallelogram  are  given  in  rods,  the  area  is  reduced  to  acres 
by  dividing  the  square  rods  by  160. 

2.  How  many  acres  are  there  in  a  piece  of  land  80  rods  in 
length,  and  40  in  breadth  ?     80  X  40  =  3200,  and  3200  -H  1 60  = 
20  acres,  Ans. 

3.  What  is  the  number  of  acres  in  a  piece  of  land  63  rods 
long  and  49  rods  wide  ?     Ans. 


Note  2d. — If  the  parallelogram  be  not  right-angled,  the  length 
must  be  multiplied  into  the  perpendicular  distance  between  the 
sides.  (See  fig.  1st.) 

The  dotted  line  in  FIG.  1. 

the  center  is  the  per- 
pendicular distance 
required ;  therefore, 
45  x  15  =  675,  the 
area  of  the  figure.  * 

Prob.  5.  To  find  the  area  of  triangles. 

The  area  of  any  triangle  is  just  equal  to  one  half  the  area 
of  a  square  or  parallelogram  of  the  same  height.  This  is  ob- 
vious with  regard  to  right-angled  triangles,  and  true  of  all 
others.  The  diagonal  „  2 

of  a  square  or  parallel- 
ogram divides  it  into  two 
equal  right-angled  tri- 
angles, and  since  the 
area  of  the  whole  figure 
is  found  by  multiplying 
the  length  into  the  breadth,  the  area  of  its  half,  that  is,  of 
either  of  the  triangles  into  which  it  is  divided  by  the  diagonal, 
is  found  by  multiplying  the  same  length  or  base  into  one  half 
the  breadth. 

We  have  then  the  following  rule  : 

RULE. — If  the  triangle  be  right-angled,  multiply  its  base  into 
half  its  perpendicular  height.  But  if  it  be  not  right-angled, 
drop  a  perpendicular  from  one  of  the  angles  to  the  opposite  side 


APPENDIX.  287 

or  base  ;  then  multiply  the  base  or  side  upon  which  the  perpendic- 
ular falls,  by  one  half  the  perpendicular. 

Ex.  t.  What  is  the  area  of  a  triangular  piece  of  land  whose 
base  is  40  rods,  and  whose  perpendicular  is  30  rods?  30-^-2 
=  15,  and  40x15  =  600  square  rods,  Ans. 

Note. — The  same  result  is  obtained  by  multiplying  the  per- 
pendicular by  one  half  the  base.  Thus,  40  -^2=20,  and  30  X 
20=600. 

2.  There  is  a  room  26  feet  long  and  18  feet  high.     If  a  line 
be  drawn  from  one  of  the  upper  corners  to  its  opposite  lower 
corner,  the  side  of  the  room  will  be  divided  into  two   equal 
triangles.     What  is  the  area  of  each  triangle,  and  also  of  the 
side  of  the  room  ?     Ans.  Each  triangle  contains  234  square 
feet,  and  the  side,  468. 

3.  How  many  acres  are  there  in  a  triangular  piece  of  land, 
whose  base  is  42  rods  and  whose  perpendicular  height  is  36 
rods  ?     Ans.  4fg  acres. 

Prob.  6.  Given  the  diameter  of  a  circle,  to  find  its  circum- 
ference. 

The  diameter  of  a  circle  is  to  its  circumference,  as  7  to  22  ; 
or,  as  1  to  3.141592.  Therefore, 

RULE. — Multiply  the  diameter  of  a  circle  by  3.141592,  and 
the  product  will  be  the  circumference  ;  or,  multiply  the  diameter 
by  22,  and  divide  the  product  by  7. 

Ex.  1.  What  is  the  circumference  of  a  wheel  whose  diam- 
eter is  8  feet.  8 x 22 -^-7=25.14+  feet,  Ans.  Or,  8x3.141592 
:=25.132736,  nearly  the  same  as  before. 

2.  What  is  the  circumference  of  a  circle  whose  diameter  is 
6  feet?     Ans.  18.85  feet. 

3.  What  is  the  circumference  of  a  circle  whose  diameter  is 
42 feet?     A?is.  132  feet. 

Prob.  7.  Given  the  circumference  of  a  circle,  to  find  the 
diameter. 

RULE. — Multiply  the  circumference  by  7  and  divide  the  pro- 
duct by  22  ;  or,  (what  will  produce  nearly  the  same  result,)  di- 
vide the  circumference  by  3.141592. 


APPENDIX. 

Ex.  1.  If  the  circumference  of  a  wheel  be  26  feet,  what  is  the 
diameter  ?  26  X  7 -22:=.- 8. 27 -f-  feet,  Ans. 

2.  What  is  the  diameter  of  a  wheel  whose  circumference  is 
50  feet?  Ans.  15.9+  feet. 

Prob.  8.  To  find  the  area  of  a  circle. 

RULE. — Multiply  half  the  diameter  by  half  the  circumference. 

Ex.  1.  If  the  diameter  of  a  circle  be  6  feet,  what  is  its  area? 
6  x  22  -f-  7  =  1 8.857  feet,  the  circumference.  Therefore,  1 8. 857 
-^-2  =  9.4285,  one  half  the  circumference,  and  6-^-2=3,  one  half 
the  diameter;  then,  9.4285x3=28.2855,4-  the  area  in  feet 
and  decimals. 

2.  If  the  diameter  be  20  feet,  what  is  the  area  of  the  circle? 
Ans.  3 14.285  feet? 

3.  What  is  the  area  of  a  circle  66  feet  in  circumference  ? 
Ans.  346.5 

Prob.  9.  To  find  the  superficial  area  of  a  globe. 

The  superficial  area  of  a  globe  is  four  times  that  of  a  circle 
of  the  same  diameter.  Therefore, 

RULE. — Find  the  area  of  a  circle  of  the  same  diameter  as  the 
given  globe,  and  multiply  that  area  by  4. 

Ex.  1.  What  is  the  superficial  area  of  a  globe  16  inches  in 
diameter?  16x22-^-7  =  50.2757.  50.2757-^2=25.13785, 
half  the  circumference,  16 -=-2 =8,  half  the  diameter;  there- 
fore, 25.13785x8x4=804.4+  square  inches,  the  required 
area. 

2.  If  the  earth  be  8000  miles  in  diameter,  and  25000  in  cir- 
cumference, how  many  square  miles  on  its  whole  surface.  Ans. 
200000000. 

Prob.  10.  To  find  the  solid  content  of  a  globe. 

RULE. — Find  the  superficial  area  by  the  preceding  problem, 
and  multiply  that  area  by  %  of  the  diameter  of  the  globe. 

Ex.  1.  If  the  diameter  of  the  earth  be  8000  miles,  and  the 
circumference  25000  miles,  how  many  cubic  miles  does  it 
contain. 


APPENDIX.  289 

The  superficial  area  was  found  by  the  preceding  problem  to 
be  200000000 ;  and i  the  earth's  diameter  is  8000-^6  =  1333£ ; 
therefore,  200000000  x  1333^=266666666666^  cubic  miles, 
Ans. 

Prob.  11.  To  find  the  solid  content  of  a  cylinder. 

A  cylinder  is  a  round  body  of  uniform  diameter,  and  of  any 
assignable  length.  Each  end  of  a  cylinder  is  therefore  a  circle. 

RULE. — Find  the  area  of  one  end  of  the  cylinder,  and  multi- 
ply that  area  by  its  length. 

Ex.  1 .  What  is  the  solid  content  of  a  cylinder  6  feet  in 
length  and  3  feet  in  diameter  ?  The  area  of  the  end,  as  found 
by  the  preceding  rule,  is  7.06  ;  and  7.06  X  6=42.36. 

2.  What  is  the  solid  content  of  a  log,  the  uniform  diameter 
of  which  is  2  feet,  and  the  length  30  feet?     Ans.  94.247  solid 
feet. 

3.  What  fraction  of  a  cord  is  there  in  a  log   10  feet  long, 
and  4  feet  in  diameter  ?     Ans.  ffj-  + 

Prob.  12.  In  a  stick  of  timber  of  uniform  thickness,  to  find 
what  length  will  make  a  solid  foot. 

RULE. — Find  the  area  of  one  end  in  inches,  and  divide  1728 
by  it ;  the  quotient  will  be  the  required  length  in  inches. 

Ex.  1.  If  a  stick  of  timber  be  16  inches  square,  what  length 
will  be  required  to  make  a  solid  foot  ?  16  X  16=256,  the  area 
of  one  end,  and  1728-f-256  =  6f  inches,  Ans. 

2.  How  much  length  of  plank  3  inches  thick  and  24  inches 
wide,  will  be  required  to  make  a  solid  foot  ?  Ans.  2  feet. 

Prob.  13.  To  find  the  solid  content  of  cones  and  pyramids. 

A  pyramid  or  cone  is  a  solid  whose  base  is  a  plain  surface, 
and  which  gradually  and  uniformly  tapers  till  it  comes  to  a 
point.  It  may  be  either  round,  square,  or  triangular. 

The  solid  content  of  such  figures  is  £  as  much  as  the  con- 
tent of  a  cylinder  of  the  same  length ;  therefore, 

RULE. — Multiply  the  area  of  the  base  by  one  third  of  the 
perpendicular  height. 

Ex.  1 .  What  is  the  solid  content  of  a  cone  60  feet  high, 
25 


290  APPENDIX. 

the  base  of  which  is  8  feet  in  diameter?  8x22-4-7—25.14 
circumference,  and  25.14-^-2x4=50.28,  area  of  the  base; 
and  50.28x20,  (one  third  the  perpendicular  height)  =  100.56, 
the  solid  content. 

2.  What  is  the  solid  content  of  a  pyramid,  whose  base  is  4 
feet  square,  and  whose  height  is  15  feet?     Ans.  80  solid  feet. 

3.  How  many  cubic  inches  in  a  cone   18  inches  high,  and 
12  inches  in  diameter  at  the  base  ?     Ans.  678.6. 

4.  There  is  a  pyramid  whose  base  is  30  feet  in  diameter, 
and  the  height  of  it  is  90  feet.     What  is  its  solid  content  ? 
Ans.  21205.73+  solid  feet. 

Prob.  14.  To  find  the  solidity  of  the  frustrum  of  a  cone  or 
pyramid. 

A  frustrum  is  that  part  of  a  cone  or  pyramid  that  remains 
after  a  part  has  been  cut  off  from  the  apex  or  top. 

RULE. — Find  the  diameter  of  each  end  of  the  cone  or  pyra- 
mid, and  multiply  them  together ;  then  to  the  product  add  one 
third  of  the  square  of  the  difference  of  the  diameters,  and  mul- 
tiply the  sum  by  the  decimal  .785398 ;  the  product  will  be  the 
mean  area  between  the  two  bases  ;  therefore,  multiply  this  mean 
area  by  the  length  of  the  frustrum. 

Ex.  1.  What  is  the  solid  content  of  a  frustrum  of  a  cone  30 
feet  in  length,  whose  base  measures  4  feet  in  diameter,  and 
whose  top  measures  2  feet  in  diameter,  4x2  =  8.  The  differ- 
ence of  diameters  is  4 — 2=2,  and  22=4,  and4-f-3  =  l|. 
Therefore,  8+ 1£  X  .785398=7.33,  nearly,  the  mean  diameter ; 
and  7.33x30=219.9,  the  required  content. 

2.  What  is  the  content  of  a  stick  of  timber,  of  which  the 
larger  end  measures  2  feet  in  diameter,  and  the  smaller  end, 
1  foot,  the  length  being  40  feet  ?  Ans.  73.3  feet,  nearly. 

• 

Prob.  15.  Given  the  length  and  diameter  of  a  log  or  stick  of 
round  timber,  to  determine  the  solid  content  of  the  greatest 
square  timber  that  can  be  obtained  from  it. 

The  area  of  the  greatest  square  that  can  be  described  upon 
the  end  of  the  given  log  must  first  be  determined  ;  that  is,  the 
area  of  a  square  inscribed  in  a  circle,  the  diameter  of  which 
equals  the  diameter  of  the  log.  How  then  is  the  area  of  an 
inscribed  square,  that  is,  of  a  square  the  corners  of  which  all 


APPENDIX. 


291 


E 


terminate  in  the  circumference  of  a  circle,  to  be  found,  the 
diameter  of  that  circle  being  given  ? 

If  the  scholar  will  examine 
the  adjoining  figure,  he  will  see 
that  the  square  ABCD,  which  is 
circumscribed  about  the  circle, 
is  equal  to  the  square  of  the  diam- 
eter of  the  circle,  since  the  diam- 
eter EN  equals  the  side  of  AD ; 
and  AD  squared  gives  the  area 
of  the  square  ABCD  :  also,  that 
the  inscribed  square  PEON,  is 
just  one  half  of  the  circumscribed 
square,  since  each  of  the  triangles 
EaO,  OaN,  NaP,  and  PaE,  into  which  the  inscribed  square 
is  divided,  is  precisely  one  half  of  each  of  the  four  squares, 
APaE,  EaOB,  OaNE,  and  NaPD,  into  which  the  circum- 
scribed square,  ABCD,  is  divided.  If  therefore  a  fourth  part  of 
the  inscribed  square  equals  one  half  of  a  fourth  part  of  the 
circumscribed  square,  the  whole  inscribed  square  equals  one 
half  of  the  circumscribed  square.  Hence, 

RULE. — Square  the  diameter  of  the  small  end  of  the  log  and 
multiply  one  half  of  this  square  by  the  length  of  the  log ;  the 
product  will  be  the  content  in  solid  feet,  if  the  length  and  diam- 
eter are  both  given  in  feet ;  but  if  one  of  them  be  expressed  in 
inches,  divide  this  product  by  144;  and  if  both,  by  1728,  and 
the  quotient  will  express  the  solid  content  in  feet. 

Ex.  1.  How  many  solid  feet  of  square  timber  may  be 
sawn  from  a  log  2  feet  in  diameter,  and  20  feet  in  length  ? 
22=4,  and  4^-2=2;  then,  2x20  =  40  solid  feet,  Ans. 

2.  How  many  solid  feet  of  square  timber  may  be  obtained 
from  a  log  10  inches  in  diameter,  and  15  feet  long?     Ans. 

solid  feet. 

3.  How  many  square  feet  of  timber  are  there  in  a  log  43 
feet  long,  allowing  the  diameter  to  be   1   foot  6  inches  at  the 
small  end;  and  how  many  solid  feet  of  slab  would  be  taken  off 
in  squaring  it,   allowing  the  diameter  at  the  large  end  to  be  2 
feet  ?     Ans.  48.37+  solid  feet  of  timber,  and  55.76  solid  feet 
of  slab. 


292  APPENDIX. 

Prob.  16.  To  find  the  capacity  of  casks  in  gallons,  &c. 

RULE. — Take  the  length  of  the  cask  between  the  heads,  and 
also  its  interior  bung  and  head  diameter,  in  inches.  Multiply 
the  difference  between  the  head  and  bung  diameter  by  .62,  when 
the  staves  are  of  ordinary  curvature  ;  but  by  .66,  if  the  curva- 
ture be  more  than  ordinary,  and  by  .58,  if  they  be  curved  less 
than  usual,  and  add  the  product  to  the  head  diameter;  the  sum  will 
be  the  mean  diameter.  Multiply  the  square  of  the  mean  diam- 
eter by  the  length,  and  this  product  by  34  for  wine,  and  by  28 

for  beer  measure,  and  from  the  product  cut  off  four  decimals. 
The  number  obtained  will  be  the  gallons,  and  the  decimal  of  a 

gallon,  the  cask  will  contain. 

Ex.  1.  How  many  gallons  of  wine  are  there  in  a  cask  whose 
bung  diameter  is  34  inches,  head  diameter  30  inches,  and 
whose  length  is  60  inches  ? 

Operation:  34  —  30=4,  the  difference  between  the  head 
and  bung  diameter ;  then, 4  X. 62=2.48,  and  2.48  +  30=32.48, 
mean  diameter,  and  32.48  x  32.48  =  1 054.9504,  square  of  mean 
diameter.  1054.9504  X  60  =  63297.0240  x  34=215.20988160, 
or  nearly  215.2  gallons,  Ans. 

2.  Suppose  the  bung  diameter  of  a  cask  of  wine  to  be  24 
inches,  and  the  head  diameter  to  be  20  inches,  and  the  length 
40  inches  ;  how  many  gallons  will  it  contain  ?  Ans.  68|  gal- 
lons, nearly. 

Prob.  17.  To  find  the  tonnage  of  double  and  single  decked 
vessels. 

The  following  is  the  rule  established  by  Congress  : 

RULE. — "  If  the  vessel  be  double-decked,  take  the  length 
thereof  from  the  fore  part  of  the  main  stern  to  the  after  part  of 
the  stern-post,  above  the  upper  deck ;  the  breadth  thereof  at  the 
broadest  part  above  the  main  wales,  half  of  which  breadth  shall  be 
accounted  the  depth  of  such  vessel,  and  then  deduct  from  the 
length,  three  fifths  of  the  breadth.  Multiply  the  remainder  by 
the  breadth,  and  the  product  by  the  depth,  and  divide  this  last 
product  by  95;  the  quotient  whereof  shall  be  deemed  the  true 
contents  or  tonnage  of  such  ship  or  vessel.  But  if  such  ship 


APPENDIX.  293 

or  vessel  be  single-decked ,  take  the  length  and  breadth,  as  above 
directed ,  deduct  from  said  length  three  fifths  of  the  breadth, 
and  take  the  depth  from  the  under  side  of  the  deck  plank  to  the 
ceiling  in  the  hold;  then  multiply  and  divide  as  aforesaid,  and  the 
quotient  shall  be  deemed  the  tonnage." 

Note. — Carpenters  multiply  together  the  length  of  the  keel, 
the  breadth  at  the  main-beam,  and  the  depth  of  the  hold,  each 
dimension  being  taken  in  feet ;  and  divide  the  product  by  95, 
for  the  tonnage  of  single-decked  vessels.  But  for  double- 
decked  vessels,  they  take  half  the  breadth  at  the  beam  for  the 
depth  of  the  hold,  and  then  work  as  before. 

Ex.  1.  Suppose  the  length  of  a  double-decked  vessel  to  be 
80  feet,  and  its  breadth  to  be  30  feet.  What  is  its  tonnage  ? 

Operation  :  30^2  =  15,  the  depth  of  the  vessel;  f  of  30  = 
18,  and  80  — 18  =  62;  and  62X30  =  1860;  then,  1860x15 
=27900  ;  and,  lastly,  27900-^-  95 =293f|  tons,  Ans. 

2.  Required  the  tonnage  of  a  double-decked  vessel,  whose 
length  is  95  feet,  and  whose  breadth  is  35  feet  1    Ans.  477T2g-. 

3.  What  is  the  tonnage  of  a  single-decked  vessel,  which  is 
75  feet  in  length,  30  feet  in  breadth,  and   10  feet  in  depth  ? 
Ans.  180  tons. 

Note. — The  tonnage  of  a  ship  of  war  is  found  by  dividing 
the  continued  product  of  the  length,  breadth,  and  depth  by  100. 

4.  What  is  the  tonnage  of  a  sloop  of  war,  whose  length  is 
96  feet,  breadth  32  feet,  and  depth  15  feet  1     Ans.  460.8  tons. 

Prob.  18.  The  difference  of  longitude  between  two  places 
being  given,  to  find  the  difference  in  time. 

As  the  circumference  of  the  earth  is  divided  into  360  degrees, 
and  as  the  sun  appears  to  pass  round  the  earth  in  24  hours, 
it  is  evident  that  360—24=15  degrees  of  the  earth's  circum- 
ference must  pass  under  the  sun  every  hour,  and  consequently 
1  degree  every  4  minutes,  and  I1  every  4  seconds.  We  have 
then  the  following  general  rule  : 

RULE. — If  the  given  degrees  be  more  than   15,  divide  them 
byl5;  the  quotient  will  be  the  hours.     But  if  the  degrees  be  less 
than  15,  multiply  them  by  4,  and  the  product  will  be  minutes. 
25* 


294 


APPENDIX. 


Multiply  also  the  given  minutes  of  motion  by  4,  and  the  product 
will  be  seconds  of  time. 

Ex.1.  Warsaw  is  situated  about  17°  east  longitude  from 
Brussels.  What  is  the  difference  in  time?  17 -^15  =  1^ 
hours,  or  1  hour,  8  minutes. 

It  is  obvious  that  the  sun  will  rise  first  at  the  most  eastern 
of  any  two  places.  Hence,  when  the  time  of  one  place  is 
given,  and  it  is  required  to  find  the  time  of  some  other  place, 
the  difference  of  time  must  be  added  to  the  given  time,  when  that 
place  is  situated  to  the  east,  and  subtracted  when  it  is  situated  to 
the  west  of  the  place,  at  which  the  time  is  given. 

2.  Turin  is  situated  8°  east  from  London,  and  Moscow  38° 
east  from  the  same  place.     What  is  the  hour  at  Moscow,  when 
it  is  9  o'clock  at  Turin  ?     Ans.  1 1  o'clock. 

3.  When  it  is  9  o'clock  at  Philadelphia,  what  time  is  it  at 
Council  Bluff,  21°  west  of  Philadelphia  ?     Ans.  36  minutes 
past  7  o'clock. 

4.  What  is  the  difference  of  time  between  Washington  and 
Harmony,  a  missionary  station  west  of  Missouri,  the  difference 
of  longitude  being  18°?     Ans.  1  hour,  12  minutes. 

Prob.  19.  The  difference  in  time  between  any  two  places 
given,  to  find  the  difference  of  longitude. 

This  is  the  reverse  of  the  preceding ;  therefore, 

RULE. — Multiply  the  hours  by  15  and  the  products  will  be  de- 

Sees.     If  minutes  and  seconds  of  time  be  given,  divide  them 
4  ;  the  quotient  of  the  minutes  will  be  degrees,  and  that  of 
$  seconds,  will  be  minutes  of  motion. 

Ex.  1 .  If  the  difference  of  time  between  two  places  be  3 
hours  and  32  minutes,  what  is  the  difference  of  longitude  1 
3  X  15=r450,  and  32^-4  =  8°  ;  and  45°+8°=:530,  Ans. 

2.  What  is  the  difference  of  longitude  between  two  places, 
if  the  time  of  one  place  be  5  hours,  15  minutes  in  advance  of 
the  time  of  the  other  place  ?  Ans.  78°  45'. 


APPENDIX.  295 


MECHANICAL    POWERS. 

When  one  body  is  made  to  communicate  motion  to  another, 
the  body  communicating  the  motion  is  called  a  power,  and  the 
one  to  which  the  motion  is  communicated,  is  called  a  weight. 
The  various  instruments  employed  to  increase  the  effect  of  a 
given  power,  are  called  Mechanical  Powers. 

The  mechanical  powers  are  six  in  number,  viz.  the  Lever, 
the  Pulley,  the  Wheel  and  Axle,  the  Inclined  Plane,  the  Screw, 
and  the  Wedge. 

The  lever  is  a  bar,  mov-  FIG.  1. 

able  about   a   fixed   point.  3  9 

It  is  represented  at  Fig.  1. 
P   represents    the  power, 
and  W  the  weight,  and  3        t^ 
and    9    the     comparative  wJB 
length  of  the  arms  of  the 
lever. 

When  the  power  and  weight  are  such  that  the  products  ob- 
tained by  multiplying  each  into  the  length  of  the  arm  to  whicn  it 
is  attached,  are  equal,  they  will  remain  in  equilibrium,  that  is, 
they  will  balance  each  other. 

In  fig.  1,  the  arm  to  which  the  power  is  applied  is  three  times 
as  long  as  that  to  which  the  weight  is  applied.  Consequently, 
the  power  is  required  to  be  only  one  third  as  heavy  as  the 
weight,  in  order  that  they  reciprocally  balance  each  other. 
We  hence  perceive  that  the.power  is  to  the  weight,  as  the  arm  to 
which  the  weight  is  applied,  is  to  that  to  which  the  power  is 
applied. 

Prob.  20.  Given  the  two  arms  of  a  lever,  and  the  power,  to 
find  what  weight  that  power  will  balance. 


i 


RULE. — Multiply  the  length  of  the  arm  to  which  the  power  is 
applied,  by  the  power,  and  divide  the  product  by  the  other  arm. 

Ex.  1.  There  is  a  lever  16  feet  in  length  divided  by  the  ful- 
crum or  point  of  support  into  two  arms,  one  of  which  is  12 


296  APPENDIX. 

feet,  and  the  other  4  feet  in  length.  What  weight  on  the  short 
arm  will  balance  20  Ib.  suspended  at  the  end  of  the  long  arm? 
20  X  12=240,  and  240^-4  =  60  Ib.  Ans. 

2.  The  arms  of  a  lever  are,  the  one  30  feet,  and  the  other  3 
feet  in  length.     What  weight  will  a  power  of  160  Ib.  at  the  ex- 
tremity of  the  long  arm,  balance  at  the  extremity  of  the  short 
arm?     Ans.  16001b. 

3.  How  many  Ib.  will  a  power  of  9  Ib.  placed  15  feet  from 
the  fulcrum  of  a  lever,  support  at  the  extremity  of  the  other 
arm,  2  feet  in  length  ?     Ans.  67^  Ib. 

Prob.  21.  Given  the  arms  of  the  lever  and  the  weight,  to  find 
the  power. 

RULE. — Multiply  the  weight  by  the  length  of  the  arm  to 
which  it  is  suspended,  and  divide  the  product  by  the  other  arm. 
The  quotient  will  be  the  power  required. 

Ex.  1 .  A  weight  of  80  Ib.  is  attached  to  an  arm  of  a  lever  3 
feet  in  length  ;  what  power  must  be  applied  to  the  other  arm, 
27  feet  in  length,  to  balance  it?  80x3=240,  and  240-^-27 
=8|lb.  Ans. 

2.  A  weight  of  20  tons  is  suspended  to  an  arm  of  a  lever  6 
inches  in  length.  What  weight  at  the  extremity  of  the  other 
arm,  40  feet  in  length,  will  balance  the  same  ?  Ans.  5  cwt. 

Prob.  22.  Given  the  power,  the  weight,  and  the  length  of  the 
lever,  to  find  the  fulcrum. 

RULE. — Add  together  the  numbers  representing  the  power 
and  weight,  and  say,  as  their  sum  is  to  the  whole  lever,  so  is  the 
number  representing  the  power  to  the  length  of  the  arm  to  which 
the  weight  is  to  be  applied ;  or,  so  is  that  representing  the  weight 
to  the  length  of  the  arm  to  which  the  power  is  to  be  applied. 

Ex.  1.  Where  must  be  the  fulcrum  in  a  lever  16  feet  in 
length,  so  that  a  power  of  20  Ib.  shall  balance  a  weight  of  60 
Ib?  20+60=80.  Therefore,  80  :  60  : :  16  :  12,  Ans.  or  length 
of  the  arm  to  which  the  power  is  to  be  applied,  and  16  — 12 
=4,  length  of  the  other  arm. 

2.  If  the  power  be  72  Ib.,  the  weight  1728  Ib.,  and  the 
lever  36  feet,  where  must  be  the  fulcrum,  in  order  that  they 
shall  balance  each  other  ?  Ans.  If-f  feet  from  the  weight. 


APPENDIX.  297 

3.  If  the  length  of  a  lever  be  10  feet,  the  power  170  lb.,  and 
the  weight  to  be  balanced  1530  lb.,  where  must  be  the  fulcrum? 
Ans.  9  feet  from  the  power. 


PULLEY. 

A  pulley  consists  of  a  wheel  movable  about  FIG.  2. 
its  axis,  and  so  arranged  as  to  be  put  in  motion 
by  means  of  a  cord  passing  over  it.  The  movable 
pulley  is  the  only  one  by  which  a  gain  in  power 
is  effected.  A  double  movable  pulley  is  repre- 
sented at  fig.  2.  By  such  a  pulley  an  equilibrium 
is  produced,  when  the  power  is  to  the  weight, 
as  one  to  the  number  of  ropes  sustaining  the 
weight.  If  a  single  movable  pulley  be  employed, 
the  weight  is  sustained  by  two  ropes ;  and  the 
power  will  be  to  the  weight,  as  1  to  2.  If  two 
movable  pulleys  be  employed,  the  weight  is  sus- 
tained by  four  ropes,  and  the  power  will  be  to 
the  weight,  as  1  to  4,  &c. ;  that  is,  each  movable 
pulleyis  sustained  by  two  ropes,  and  consequently 
doubles  the  effect  of  the  power  employed. 

Prob.  23.  Given  the  number  of  movable  pul- 
leys and  the  power,  to  find  the  weight. 

RULE. — Multiply  the  power  applied  by  twice  the  number  of 
movable  pulleys. 

Ex.  1.  If  in  a  system  of  pulleys  there  be  three  movable 
pulleys,  what  weight  will  a  power  of  72  lb.  balance  ?  72  x 
6  =  432lb.  Ans. 

2.  What  weight  will  a  power  of  15  lb.  balance  by  a  system 
of  6  movable  pulleys  ?     Ans.  180  lb. 

3.  By  the  aid  of  a  system  of   12  movable  pulleys,  how 
many  pounds  will  a  man  sustain  who  is  capable  of  applying  a 
power  of  1501b.?     Ans.  3600. 

Prob.  24.  Given  the  number  of  movable  pulleys,  and  the 
weight  to  be  balanced,  to  find  the  required  power. 


298  APPENDIX. 

RULE. — Divide  the  weight  by  twice  the  number  of  movable 
pulleys. 

Ex.  1.  How  many  pounds  would  be  required  by  the  aid  of 
two  movable  pulleys,  to  sustain  800  Ib.  ?     Ans.  200  Ib. 

2.  By  the  aid  of  10  movable  pulleys,  how  many  Ib.  would 
be  required  to  balance  2000  Ib.  ?     Ans.  100  Ib. 

3.  What  weight  would  balance  10  pounds  of  silver  by  the 
aid  of  a  system  of  100  pulleys  1     Ans.  12  pwt.  • 


WHEEL  AND  AXLE. 

The  wheel  and  axle  form  a  FIG.  3. 

kind  of  perpetual  lever,  the 
long  arm  of  which  is  the  semi- 
diameter  of  the  wheel,  and  the 
short  arm,  the  semi-diameter  of 
the  axle.  Consequently,  the 
power  of  the  wheel  and  axle 
is  increased  either  by  making 
the  wheel  larger,  while  the  axle 
remains  unaltered,  or  by  making 
the  axle  smaller,  while  the 
wheel  remains  the  same.  The 
power  must  always  be  applied 
to  the  wheel,  and  the  weight  to 
the  axle,  when  an  increase  of 
action  is  to  be  gained.  In  the 
use  of  the  wheel  and  axle,  an  equilibrium  is  produced,  when 
the  power  is  to  the  weight  as  the  semi-diameter  of  the  axle  is  to 
the  semi-diameter  of  the  wheel. 

Prob.  25.  Given  the  diameter  of  the  wheel,  the  diameter  of 
the  axle,  and  the  power,  to  find  the  weight. 

RULE. — Multiply  the  diameter  of  the  ivheel  by  the  power  ap- 
plied, and  divide  the  product  by  the  diameter  of  the  axle. 

Ex.  1 .  If  the  diameter  of  the  axle  be  six  inches,  and  that  of 
the  wheel  6  feet,  what  weight  attached  to  the  axle,  will  16 
Ib.  attached  to  the  wheel,  balance  ?  Ans.  192  Ib. 


APPENDIX.  299 

2.  If  the  diameter  of  a  wheel  be  20  feet,  and  that  of  the 
axle  2  feet,  and  a  power  of  400  Ib.  be  applied  to  the  wheel, 
what  weight  will  be  balanced  at  the  axle  ?  Ans.  4000  Ib. 

Prob.  26.  Given  the  diameter  of  the  wheel,  the  diameter 
of  the  axle,  and  the  weight  to  be  balanced,  to  find  the  required 
power. 

RULE. — Multiply  the  diameter  of  the  axle  by  the  weight,  and 
divide  the  product  by  the  diameter  of  the  wheel. 

Ex.  1 .  If  the  diameter  of  the  axle  be  6  inches,  and  the  diam- 
eter of  the  wheel  12  feet,  what  power  will  balance  a  weight 
of  360  Ib?  Ans.  15lb. 

2.  If  the  diameter  of  the  axle  be  2  feet,  and  the  diameter  of 
the  wheel  20  feet,  what  power  will  balance  a  weight  of  4000 
Ib.  ?  Ans.  400  Ib. 


THE     INCLINED     PLANE. 

An  inclined  plane  is  a  plane  that  makes  with  the  ground  or 
floor  some  certain  angle  less  than  a  right  angle.  In  the  appli- 
cation of  this  instrument,  the  ratio  of  the  power  and  weight  is 
always  the  same  as  that  of  the  height  and  length  of  the  plane. 

Prob.  27.  Given  the  length  and  height  of  the  plane,  and  also 
the  power,  to  determine  the  weight. 

RULE. — Multiply  the  power  by  the  length  of  the  plane,  and 
ivide  the  product  by  its  perpendicular  height. 

Ex.  1.  If  the  length  of  a  plane  be  16  feet,  and  its  height  4 
feet,  what  weight  will  a  power  of  32  Ib.  sustain  1  Ans.  1281b. 

2.  What  weight  will  164  Ib.  sustain  on  a  plane  112  feet  in 
length  and  3  feet  in  height  ?  Ans.  6122|  Ib. 

Prob.  28.  Given  the  length  and  height  of  the  plane,  and  also 
the  weight,  to  find  what  power  will  sustain  it. 

RULE. — Multiply  the  weight  by  the  height  of  the  plane,  and 
divide  the  product  by  the  length. 


300  APPENDIX. 

Ex.  1.  What  power  will  balance  J28lb.  on  an  inclined  plane, 
the  length  of  which  is  16  feet,  and  the  height  4  feet?  Ans. 
32  Ib. 

2.  Suppose  on  a  railroad  there  is  an  inclined  plane  150 
rods  in  length,  and  rising  to  the  perpendicular  height  of  50  feet ; 
what  power  will  be  required  to  sustain  a  weight  of  84000  Ib. 
Ans.  28000  Ib. 


THE    WEDGE. 

The  wedge  is  composed  of  two  inclined  planes,  whose  bases 
unite  and  form  the  base  of  the  wedge.  The  power  applied  to 
the  wedge  is  to  the  effect  produced  at  the  side  of  the  wedge, 
as  the  thickness  of  the  head  to  the  length  of  the  wedge,  no  allow- 
ance being  made  for  friction. 

Prob.  29.  Given  the  thickness  of  the  head,  the  length  of 
the  side,  and  the  power  acting  upon  the  head  of  a  wedge,  to 
determine  the  force  produced  on  the  side. 

RULE. — Multiply  the  length  of  the  wedge  by  the  power  ap- 
plied^ and  divide  the  product  by  the  thickness  of  the  head. 

Ex.  1.  If  the  length  of  a  wedge  be  12  inches,  the  thickness 
of  the  head  3  inches,  and  the  force  applied,  64  Ib.,  what  will 
be  the  resistance  at  the  side  1  Ans.  256  Ib. 

2.  If  the  length  of  a  wedge  be  -32  inches,  the  thickness  of 
the  head  2  inches,  and  the  force  applied  16001b.,  what  will 
be  the  resistance  at  the  side  ?  Ans.  25600  Ib. 

Prob.  30.  Given  the  length  of  the  side,  the  thickness  of  the 
head,  and  the  resistance  upon  the  side  of  a  wedge,  to  find  the 
force  acting  upon  the  head. 

RULE. — Multiply  the  resistance  at  the  side  by  the  thickness 
of  the  head,  and  divide  the  product  by  the  length  of  the  side  of 
the  wedge. 

Ex.  1.  If  the  length  of  the  wedge  be  32  inches,  the  thick- 
ness of  the  head  2  inches,  and  the  resistance  at  the  side  be 
25600  Ib.  what  must  be  the  force  upon  the  head,  no  allowance 
being  made  for  friction  ?  Ans.  1600  Ib. 


APPENDIX. 


301 


2.  If  the  resistance  at  the  side  of  a  wedge  be  20000  lb., 
the  length  of  the  wedge  20  inches,  and  the  thickness  of  the 
head  3  inches,  what  force  is  required  to  be  applied  to  counter- 
act the  resistance  at  the  sides  ?  Ans.  3000  lb. 


THE    SCREW. 

The  screw  is  a  cylindrical  piece  of  wood,  or  metal,  with  a 
spiral  thread  running  round  it  with  a  gradual  and  uniform  in- 
clination. The  thread,  therefore,  forms  an  inclined  plane  of 
the  same  length  as  the  thread  itself,  and  whose  height  equals 
the  length  of  the  screw. 

It  is  then  obvious  that  the  power  of  the  screw  depends  in 
part  upon  the  fineness  of  the  threads,  since  the  finer  they  are, 
the  greater  will  be  the  length  of  the  plane,  while  its  height 
remains  the  same.     The  power  of  the 
screw  also  depends  in  part  on  the  FIG.  4. 

length  of  the  lever  employed  to  move 
it.  It  is,  therefore,  a  compound  power, 
involving  in  its  action  both  the  princi- 
ple of  the  lever  and  the  inclined  plane, 
and  is  consequently  more  efficient'than 
either  of  the  other  mechanical  powers. 
In  the  action  of  this  instrument,  an 
equilibrium  between  the  power  and 
weight  is  produced,  when  the  power 
is  to  the  weight  as  the  distance  between 
two  contiguous  threads  of  the  screw  is 
to  the  circumference  of  the  circle  de- 
scribed by  the  power  in  one  revolution. 

Prob.  31.  Given  the  distance  between  the  threads  of  the 
screw,  the  length  of  the  lever,  and  the  power  applied  to  find 
the  weight. 

RULE. — Multiply  the  circumference  of  the  circle  described  by 
one  revolution  of  the  lever  by  the  power  applied,  and  divide  the 
product  by  the  distance  between  the  threads  of  the  screw. 

Note. — The  lever  is  the  semi-diameter  of  the  circle  it  de- 
scribes ;   the  circumference  of  the  circle  may  therefore  be 
found  by  Prob.  6th. 
26 


302  APPENDIX. 

Ex.  1.  If  the  threads  of  a  screw  be  2  inches  apart,  the 
lever  40  inches  in  length,  and  a  power  of  60  Ib.  be  applied  to 
the  end  of  the  lever,  what  weight  will  be  required  to  produce 
an  equilibrium?  Ans.  7542  Ib.  13  oz.  11|-  drams. 

2.  If  the  distance  between  the  threads  of  a  screw  be  1  inch, 
the  length  of  the  lever  16  inches,  and  the  power  applied  to  the 
end  of  the  lever  be  40  Ib.,  what  weight  will  produce  an  equi- 
librium ?  Ans.  4022.8+  Ib. 

Prob.  32.  Given  the  weight,  the  length  of  the  lever,  and 
the  distance  between  the  threads  of  the  screw,  to  find  the  pow- 
er required  to  produce  an  equilibrium. 

RULE. — Multiply  the  given  weight  by  the  distance  between 
the  threads  of  the  screw,  and  divide  the  product  by  the  circum- 
ference of  the  circle  described  by  one  revolution  of  the  lever. 

Ex.  1.  How  many  pounds  applied  to  the  end  of  a  lever 
16  inches  long,  will  balance  4022.8  Ib.  upon  a  screw,  whose 
threads  are  one  inch  apart  ?  Ans.  40  Ib. 

2.  How  many  pounds  applied  to  the  end  of  a  lever  30  inches 
long,  will  produce  an  equilibrium  with  4000  Ib.  attached  to 
a  screw  whose  threads  are  2  inches  apart?  Ans.  42?  Ib. 
nearly. 


STEAM    POWER. 

The  power  of  the  steam  engine  is  usually  estimated  by  com- 
paring it  with  that  of  the  horse.  There  are,  however,  various 
circumstances,  which  tend  to  render  the  power  of  the  horse  an 
indefinite  standard  ;  such  as  the  natural  strength  of  the  ani- 
mal,— the  training  it  may  have  received  to  enable  it  to  exert  its 
strength  to  the  best  advantage, — its  position  while  exerting  its 
strength, — the  length  of  time  the  exertion  is  to  be  continued, 
and  the  velocity  of  motion  required.  Before  the  term,  "  horse 
power,"  could  be  employed  to  communicate  any  uniform  idea 
of  force  exerted,  it  was  therefore  necessary  to  determine  by 
experiment,  what  was  the  medium  power  of  the  horse,  making 
due  allowance  for  varying  circumstances. 

Since,  however,  the  power  required  to  move  any  given  body 
in  a  horizontal  direction  is  dependent  on  various  circumstances 
besides  its  actual  weight,  and  as  a  diminished  force  only  is  re- 


APPENDIX.  303 

quired  to  continue  a  body  in  motion,  when  motion  in  that  direc- 
tion is  once  communicated,  a  medium  power  could  be  ob- 
tained only  by  repeated  experiments  on  the  power  of  the 
horse  in  raising  weights  in  a  perpendicular  direction,  like 
raising  a  stone  from  a  well,  while  the  animal  was  made  to  act 
in  a  horizontal  direction  through  the  agency  of  a  single  pulley. 
Experiments  were  accordingly  instituted,  the  result  of  which 
has  been  to  establish  the  fact,  that  a  one  horse  power  is  a  power 
competent  to  raise  2250  Ib.  one  mile  in  a  perpendicular  direc- 
tion in  6  hours  of  time.  This  fact  being  thus  established,  the 
term,  "  horse  power,"  has  acquired  a  distinctness  of  meaning, 
which  renders  it  a  convenient  standard  by  which  to  estimate 
the  power  of  machinery,  especially  when  propelled  by  steam. 
The  weight  given,  viz.  2250 Ib.,  is  a  little  more  than  a  ton  ;  an 
amount  entirely  beyond  the  power  of  any  horse  to  move  as  above 
stated.  The  velocity  of  one  mile  in  6  hours  of  time,  is,  on 
the  other  hand,  far  less  than  the  natural  velocity  of  the  horse. 
How  are  we  then  to  understand  the  estimation  of  horse  power 
here  presented  ?  It  will  be  remembered  that  in  all  machinery 
by  which  a  gain  of  power  is  effected,  the  power  applied  to  put 
the  machinery  in  motion,  always  moves  through  a  greater  space 
than  the  body  moved  by  the  same  machinery  ;  or,  in  other 
words,  "  what  is  gained  in  power  is  lost  in  time."  This 
fact  presents  us  with  a  full  explanation  of  the  matter.  A  horse 
with  a  moderate  burden  will  easily  move  3  miles  per  hour, 
and  in  6  hours,  6x3  =  18  miles.  Now,  suppose  the  horse  to 
act  through  machinery  so  constructed,  that  while  he  moves  at 
the  rate  of  18  miles  in  6  hours,  the  weight  attached  to  and 
acted  upon  by  the  same  machinery,  moves  only  one  mile  ;  how 
great  a  power  would  the  horse  be  required  to  exert  to  raise 
2250  Ib.  that  distance  ?  Evidently,  a  power  equal  to  -^  of 
2250  Ib.  or  125  Ib.,  a  weight  entirely  within  the  power  of  a 
horse  of  ordinary  strength  to  raise  at  the  velocity  proposed. 
This  reduces  the  statement  of  a  single  horse  power  to  a  form 
easy  to  be  comprehended ;  viz.  it  is  a  power  competent  to  raise 
125  Ib.  18  miles  in  6  hours,  or  3  miles  per  hour. 

Prob.  33.  To  determine  the  horse  power  required  to  raise  a 
given  weight  any  specified  distance  per  hour. 

RULE. — Multiply  the  weight  reduced  to  pounds  by  the  dis- 
tance through  which  it  is  to  be  moved,  and  make  the  product  a 
dividend.  Then,  having  reduced  3  miles  to  the  same  denomina- 


304  APPENDIX. 

tion  with  the  given  distance,  multiply  it  by  125,  and  make  the 
product  a  divisor.  Divide,  and  the  quotient  will  express  the  re- 
quired horse  power. 

Ex.  1 .  What  amount  of  horse  power  will  be  required  to  raise 
14  T.  14  cwt.  2  qr.  16  lb.,  60  feet  per  hour  ? 

Operation:  14  T.  14  cwt.  2  qr.  161b.=33000  lb.;  and  33000 
X 60 =1980000,  the  dividend.  3  miles  =  15 840  feet;  and 
15840 X  125  =  1980000,  the  divisor;  then,  1980000^-1980000 
=  1,  Ans. 

Note. — If  the  time  given  be  less  than  one  hour,  the  required 
power  must  be  proportionally  greater.  Had  the  time  in  the 
preceding  sum  been  20  minutes,  instead  of  one  hour,  that  is, 
£  of  the  time  given,  three  times  as  much  power  would  have  been 
required.  Hence,  the  power  required  is  found  by  multiplying  1 
by  the  direct  ratio  of  125  and  the  given  weight,  and  by  the  in- 
verse ratio  of  1  hour  and  the  given  time. 

2.  What  amount  of  horse  power  is  necessary  to  raise  60 
tons  3  feet  per  minute  1  Ans.  12.2+  horse  power. 

Prob.  34.  Given  the  horse  power,  the  distance,  and  the  time, 
to  determine  the  weight  that  may  be  raised. 

RULE. — Multiply  125  by  the  given  horse  power ;  the  product 
will  be  the  weight  that  power  will  raise  3  miles  per  hour.  Mul- 
tiply this  by  3  miles  reduced  to  the  same  denomination  as  the 
given  distance,  and  divide  the  product  by  that  distance,  and  the 
number  obtained  will  determine  the  pounds  that  may  be  raised 
the  given  distance  in  one  hour ;  then,  lastly,  multiply  by  the 
given  time  in  minutes,  and  divide  the  product  by  60;  the  quotient 
will  be  the  weight  required. 

Ex.  1 .  What  weight  will  a  5  horse  power  raise  120  feet  in  20 
minutes  of  time  ? 

Solution:  125x5=625  ;  then,  since  3  miles  equals  15840 
feet,  625x15840  =  9900000;  and  9900000 -f- 120  =  82500  ; 
and  82500  x  20-^60=27500  lb.,  Ans. 

2.  What  weight  will  an  8  horse  power  raise  360  feet  in  90 
minutes  of  time  ?  Ans.  29  T.  9  cwt.  1  qr.  4  lb. 


APPENDIX. 


305 


PROBLEMS   IN   INTEREST. 

Prob.  35.  Time,  rate  per  cent.,  and  amount  given,  to  find  the 
principal. 

RULE. — Divide  the  given  amount  by  the  amount  of  $1  or 
I  £.  for  the  given  rate  and  time. 

Ex.  1 .  What  sum  of  money  will  amount  to  $360,  at  6  per 
cent.,  in  3  years  and  4  months  ? 

The  amount  of  $1  for  3  years  and  4  months,  is  $1.20  ;  and 
$360-h$1.20  =  $300,  Ans. 

2.  What  sum  of  money  will,  in  4  years,  6  months,  at  6  per 
cent,  per  annum,  amount  to  $1016  ?  Ans.  $800. 

Prob.  36.  Time,  rate  per  cent.,  and  interest  given,  to  find  the 
principal. 

RULE. — Divide  the  given  interest  by  the  interest  of  one 
dollar  for  the  given  time  and  rate  per  cent. 

Ex.  1.  If  a  man's  annual  interest  be  $1200,  what  is  his 
capita],  the  rate  per  cent,  being  6  1  Ans.  $20000. 

2.  How  much  money  on  interest,  at  6  per  cent.,  will  gain 
$36  in  1  year  and  8  months  ?  Ans.  $360. 

Prob.  37.  Given  the  principal,  interest,  and  time,  to  find  the 
rate  per  cent. 

RULE. — Divide  the  given  interest  by  the  interest  of  the  given 
principal,  at  one  per  cent,  for  the  given  time. 

Ex.  1.  A  man  having  $4000  on  interest,  at  the  expiration 
of  one  year  received  $240  interest  on  the  same.  What  rate 
per  cent,  did  he  receive  ?  Ans.  6  per  cent. 

2.  If  $200  be  paid  as  interest  on  $2000  for  2  years  and 
6  months,  what  is  the  rate  per  cent.  ?  Ans.  4  per  cent. 

Prob.  37.  Given  the  principal,  rate  per  cent.,  and  interest, 
to  find  the  time. 

RULE. — Divide  the  given  interest  by  the  interest  of  the  prin- 
cipal for  one  year ;  the  quotient  will  be  the  time. 
26* 


306  APPENDIX. 

Ex.  1.  Paid  $108  interest  on  a  note  of  $600.  How  long 
had  the  note  been  on  interest,  the  rate  per  cent,  being  6  ?  Ans. 
3  years. 

2.  Paid  8400  interest  on  a  note  of  $800,  the  rate  per  cent, 
being  4.  How  long  had  the  note  been  on  interest  ?  Ans.  12 
years  and  6  months. 


ANNUITIES. 

An  annuity  is  a  sum  of  money  payable  annually  for  any  given 
number  of  years,  or  forever. 

When  any  sum  of  money,  payable  annually,  has  remained 
unpaid  for  any  number  of  years,  the  amount  due  is  the  sum  of 
all  the  annuities  which  are  in  arrears,  together  with  the  interest 
on  each  annuity  for  the  time  it  has  remained  unpaid. 

Prob.  38.  Given  the  annuity,  the  rate  per  cent.,  and  the  time 
of  arrears,  to  find  the  amount  due. 

RULE. — Find  the  amount  of  each  annuity  for  the  time  it  has 
remained  unpaid,  at  the  given  rate  per  cent. ;  the  sum  of  the 
amounts  thus  obtained  will  be  the  sum  required. 

Ex.  1.  A  man,  who  was  in  the  receipt  of  an  annual  pension 
of  $200,  was  required  to  forego  the  payment  four  years  in 
succession.  What  was  due  at  the  expiration  of  that  time,  the 
rate  per  cent,  being  6  1 

The  annuity  of  the  fourth  year  would  receive  no  interest, 
as  it  was  not  due  till  the  year  expired,  and  consequently 
amounted  to  only  $200.  The  annuity  of  the  third  year  was 
entitled  to  one  yearrs  interest,  and  therefore  amounted  to  $212. 
The  annuity  of  the  second  year  was  entitled,  to  two  years  in- 
terest, and  amounted  to  $224.  The  annuity  of  the  first  year 
was  entitled  to  three  years  interest,  and  amounted  to  $236. 
Therefore,  200-j-212  +  224+236=$872,  Ans. 

2.  What  is  the  value  of  an  annuity  of  $600,  which  has 
remained  unpaid  for  8  years,  interest  at  6  per  cent,  being  al- 
lowed ?  Ans.  $5808. 

Prob.  39.  Given  the  annuity  and  rate  per  cent.,  to  find  its 
present  worth  for  any  number  of  years. 


APPENDIX.  307 

RULE. — Divide  each  annuity  by  the  amount  of  $1  or  1  £. 
for  the  time  before  it  becomes  due  ;  the  quotient  will  be  the  pres- 
ent worth  of  the  several  annuities,  and  their  sum  will  be  the 
amount  required. 

Ex.  1.  What  is  the  present  worth  of  an  annuity  of  $600, 
for  three  years  to  come,  6  per  cent,  being  allowed  for  present 
payment  1 

The  present  worth  of  the  first  year  is  $600  -4-  $  1 .06  =  566.037 ; 
of  the  second  year,  $600-f-$1.12  =  $535.714 ;  and  of  the 
third  year,  $600 -^$1.1 8  =  $508.474.  Then,  $566.037  + 
$535.714  +  $508.474  =  $1610.225,  Ans. 

2.  What  is  the  present  worth  of  an  annuity  of  30  <£.  for  5 
years  to  come,  at  4  per  cent.  ?  Ans.  134  £.  5  s.  5  d.  4- 


ANNUITIES  AT  COMPOUND  INTEREST. 

The  amount  of  an  annuity  at  compound  interest,  is  obtained 
by  computing  the  compound  interest  of  the  several  payments,  and 
then  finding  their  sum.  Operations  of  this  nature  may  obvi- 
ously be  solved  by  geometrical  progression,  by  making  $1  the 
first  term  of  a  geometrical  series,  and  the  amount  of  $1  for 
one  year  at  the  given  rate  per  cent.,  the  ratio.  The  number 
of  terms  will  always  be  the  same  as  the  number  of  years, 
(See  Geometrical  Progression.) 

Ex.  1.  What  will  an  annuity  of  $60  per  annum  amount  to  in 
4  years,  at  6  per  cent.?  $1  is  the  first  term,  $1.06  the  ratio. 
Therefore, 

— O^X  60  =  $262.476,+  Ans. 

We  have  then  the  following  general  principle.  Raise  the 
ratio  (which  is  always  found  by  adding  the  per  cent,  to  $  1 )  to 
a  power  equal  to  the  number  of  years ;  from  this,  subtract  1, 
then  divide  the  remainder  by  the  ratio  less  1 ,  (that  is,  by  the  de- 
cimal part  of  the  ratio  only,)  and  multiply  the  quotient  by  the 
annuity. 


308 


APPENDIX, 


2.  What  is  the  amount  of  an  annuity  of  $400,  which  has 
remained  unpaid  for  20  years,  at  6  per  cent,  compound  inter- 
est ?  Ans.  14714.23.  + 

Or,  the  answers  to  such  sums  may  be  found  by  multiplying 
the  amount  of  one  dollar  for  the  given  number  of  years  and 
rate  percent,  as  found  in  the  following  table,  by  the  annuity  : 

TABLE 

SHOWING  THE  AMOUNT  OF  $1  AT  5  OR  6  PER  CENT.  COMPOUND 

INTEREST,  FOR  ANY  NUMBER  OF  YEARS  BETWEEN  ONE 

AND  FORTY. 


yr- 

5  per  cent. 

1.000000 

yr. 
1 

6  per  cent. 

1.000000 

yr. 

21 

5  per  cent. 

35.719252 

yr. 
21 

6  per  cent. 

39.992727 

2 

2.050000 

2 

2.060000 

22 

38.505214 

22 

43.392290 

3 

3.152500 

3 

3.183600 

23 

41.430475 

23 

46.995828 

4 

4.310125 

4 

4.374616 

24 

44.501999 

24 

50.815577 

5 

5.525631 

5 

5.637093 

25 

47.727099 

25 

54.864512 

6 

6.801913 

6 

6.975319 

26 

51.113454 

26 

59.156383 

7 

8.142008 

7 

8.393838 

27 

54.669126 

27 

63.705766 

8 

9.549109 

8 

9.897468 

28 

58.402583 

28 

68.528112 

9 

11.026564 

9 

11.491316 

29 

62.322712 

29 

73.639798 

10 

12.577893 

10 

13.180795 

30 

66.438847 

30 

79.058186 

11 

14.206787 

11 

14.971643 

31 

70.760790 

31 

84.801677 

12 

15.917127 

12 

16.869941 

32 

75.298829 

32 

90.889778 

13 

17.712983 

13 

18.882138 

33 

80.063771 

33 

97.343165 

14 

19.598632 

14 

21.015066 

34 

85.066959 

34 

104.183755 

15 

21.578564 

15 

23.275970 

35 

90.220307 

35 

111.434780 

16 

23.657492 

16 

25.672528 

36 

95.836323 

36 

119.120867 

17J25.840366 

17 

28.212880 

37 

101.628139 

37 

127.268119 

1828.132385 

18 

30.905653 

38 

107.709546 

38 

135.904206 

19 

30.539004 

19 

33.759992 

39 

114.095023 

39 

145.058458 

20 

33.065954 

20 

36.785591 

40 

120.799774 

40 

154.761966 

3.  What  is  the  amount  of  an  annuity  of  $150,  which  has 
remained  unpaid  for  12  years,  compound  interest  6  per  cent.  ? 
Ans.  $2530.49.+ 

The  amount  of  $1  for  12  years  in  the  above  table,  is  $16.- 
869941,  and  $16.869941  x  150  =  $2530.49,+  Ans. 

4.  What  is  the  worth  of  an  annual  salary  of  $1000,  which 


APPENDIX, 


309 


has  remained  unpaid  for  15  years,  compound  interest  at  6  per 
cent.  ?     Ans.  $23275.97. 

5.  What  is  the  value  of  an  annuity  of  $150,  to  continue  30 
years,  at  5  per  cent,  compound  interest  ?     Ans.  $9965. 827. -f- 

TABLE 

SHOWING  THE  PRSEENT  WORTH  OF  AN  ANNUITY  OF  $1  AT  5  OR 

6  PER  CENT.  FOR  ANY  NUMBER  OF  YEARS  BETWEEN 

ONE  AND  FORTY. 


Jt. 

2 

5  per  cent. 

0.952381 
1.859410 

yr- 

1 
o 

6  per  cent. 

0.943396 
1.833393 

yr.  1  5  per  cent. 

21  12.821153 
22113.163003 

yr. 
21 
22 

6  per  cent. 

11.764077 
12.041582 

3 

2.723248 

3  2.673012 

23 

13.488574 

23 

12.303379 

4 
5 

3.545950 
4.329477 

4 
5 

3.465106 
4.212364 

24  13.798642 
25  14.093945 

24 
25 

12.550358 
12.783356 

6 

5.075692 

6 

4.917324 

26 

14.375185 

26 

13.003166 

7 

5.786373 

7 

5.582381 

27 

14.643034 

27 

13.210534 

8 

6.463213 

8 

6.209794 

28 

14.898127 

28 

13.406164 

9 

7.107822 

9 

6.801692 

29 

15.141074 

29 

13.590721 

10 

7.721735 

10 

7.360087 

30 

15.372451 

30 

13.764831 

11 

8.306414 

ill 

7.886875 

31 

15.592810 

31 

13.929086 

12 

8.863252 

12 

8.388844 

32 

15.802677 

32 

14.084043 

13 

9.393573 

13 

8.852683 

33 

16.002549 

33 

14.230230 

14 

9.898641 

14 

9.294984 

34 

16.192904 

34 

14.368141 

15  10.379658 

15 

9.712249 

35 

16.374194 

35 

14.498246 

16  10.837770 

16 

10.105895 

36 

16.546852 

36 

14.620987 

1711.274066 
18'11.689587 

17 

18 

10.477260 
10.827603 

37  16.711287 
38  16.867893 

37 

38 

14.736780 
14.846019 

19  12.085321 
20:12.462216 

19  11.158116 
20111.469921 

39  17.017041 
40117.159086 

39 
40 

14.949075 
15.046297 

To  find  the  value  of  an  annuity  by  the  preceding  table,  mul- 
tiply the  value  of  one  dollar  as  given  in  the  preceding  table  for 
the  given  number  of  years  and  rate  per  cent,  by  the  given  num- 
ber of  dollars. 

Ex.  1.  What  sum  of  money  will  purchase  an  annuity  of 
$400,  to  continue  12  years,  at  6  per  cent,  discount? 

The  value  of  $1  for  12  years  is  $8.388844,  and  8.388844  X 
=  $3355.537,+  Ans. 


310  APPENDIX. 

2.  What  is  the  present  value  of  an  annuity  of  $1200,  to 
continue  16  years,  at  6  per  cent  discount  1  Ans.  12 127.074.  -f 

3.  How  much  must  be  paid  for  an   annuity  of  $75,  to  con- 
tinue 30 years,  at  6  per  cent,  discount?     Ans.  $1032.352.+ 


ASSESSMENT    OF   TAXES. 

A  tax  is  a  sum  of  money  collected  from  the  citizens  of  a 
State,  county,  or  town,  for  defraying  the  expenses  necessarily 
incurred  in  the  administration  of  justice,  and  in  works  of 
common  utility. 

The  sum  each  man  is  required  to  pay,  depends,  for  the  most 
part,  on  the  amount  of  his  property.  To  this  the  poll  tax  forms 
the  only  exception  ;  which  is  a  tax  required  of  every  male  in- 
habitant of  a  State,  who  has  attained  the  age  of  twenty-one 
years,  independently  of  the  property  he  may  possess. 

In  levying  a  tax  upon  any  community,  first  take  a  complete 
list  of  all  the  property  of  the  town  on  which  the  tax  is  to  be 
laid,  and  also  of  the  number  of  polls  to  be  taxed.  Next,  de- 
termine the  amount  of  the  poll  taxes  by  multiplying  the  num- 
ber of  polls  by  the  tax  on  each.  This  being  determined,  sub- 
tract it  from  the  whole  sum  to  be  raised,  and  the  remainder 
will  be  the  sum  to  be  raised  on  the  property  of  the  town,  both 
real  and  personal.  Then  to  ascertain  the  percentage,  divide 
the  tax  to  be  raised  by  the  whole  amount  of  taxable  property, 
and  the  quotient  will  be  the  tax  on  one  dollar.  To  find  what 
tax  any  individual  pays,  multiply  his  inventory  by  this  quotient, 
or  tax  per  dollar. 

Ex.  1 .  What  will  be  A's  tax,  if  he  own  real  estate  to  the 
amount  of  $1340,  and  personal  property  to  the  amount  of 
$874,  if  the  town  in  which  he  lives,  the  inventory  of  which 
is  $32265,  raise  a  tax  of  $1129.95,  there  being  270  polls 
which  are  taxed  60  cents  each,  and  for  two  of  which  he  pays  ? 

270  polls  x  60  =  $162. 00,  amount  of  the  poll  taxes.  There- 
fore, $1129.95 — $162.00rr$967.95,  the  sum  to  be  levied  on 
the  property.  Hence,  $967.95-^-  $32265 =3  cents,  the  tax  on 
one  dollar.  Then,  $1340  x  .03=r$40.20,  and  $874  x. 03  = 
$26.22,  and  two  polls  at  60  cents  each=$1.20.  Therefore, 
$40.20+$26.22+$1.20  =  67.62,  Ans. 


APPENDIX. 


311 


After  finding  what  the  tax  is  per  dollar,  a  table  may  be 
formed  like  the  following,  which  will  expedite  the  operations  : 


TABLE. 


ax  on  SI  is  .03 

Tax  on  $10  is    .30 

Tax  on   $100  js   3.00 

2 

.06 

20  "     .60 

200 

6.00 

3 

.09 

30 

.90 

300 

9.00 

4 

.12 

40 

1.20 

400 

12.00 

5 

.15 

50 

1.50 

500 

15.00 

6 

.18 

60 

1.80 

600 

18.00 

fi 

.21 

70 

2.10 

700 

21.00 

8 

'  .24 

80 

2.40 

800 

24.00 

9 

'  .27 

90 

2.70 

900 

27.00 

1000 

:  30.00 

Now,  to  determine  A's  tax  from  this  table,  $1340+ $874  = 
$2214. 

The  tax  on  $2000  =  $60.00 

200=      6.00 

"  10=        .30 

"  4=        .12 

Two  polls  at  60  ct.  =      1 .20 

$67.62    Ans.  as  before. 

2.  From  the  preceding  table,  calculate  B's  tax,  whose  real 
estate  is  valued  at  $5620,  and  his  personal  property  at  $1162, 
and  who  pays  3  poll  taxes,  each  75  cents.     Ans.  $205.71. 

3.  Form  a  table  and  calculate  the  tax  of  A.,  whose  property 
is  valued  at  $1500  ;  of  B.,  whose  property  is  $2000  ;  of  C., 
whose  property  is  $1200;  of  D.,  whose   property  is  $3000, 
and  of  E.,  whose  property  is   estimated  at  $3500  ;  supposing 
them  to  live  in  a  town  whose  inventory  is   $  1 000000,  and  on 
which  a  tax  of  $4280  is  to  be  levied,  the  number  of  polls  being 
400,  and  paying  each  -70  cents.     Ans.   A's  tax  will  be  $6  ; 
B's,  $8  ;  C's,  $4.80  ;  D's,  $12  ;  and  E's,  $14. 

4.  A  tax  of  $3000  is  to  be  raised  on  a  certain  town,  the 
inventory  of  which  is  $60000  ;  the  number  of  polls  250,  taxed 
each  75  cents.    What  is  the  amount  of  A's  tax,  whose  property 
is  valued  at  $2500,  and   who   pays   for   two   polls  ?     Ans. 
$118.68$. 


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